Factorization[class 9] – Full Chapter Exercise solutions

After studying the chapter Factorization[class 9], you learn to find the factors of simple monomial expressions, find the factors of binomial expressions, find the factors of some cubic expressions.

3.2.1 Introduction to Factorization[class 9]:

Different methods of Factorization[class 9]:

Method 1: Factorization[class 9] taking common factors:

When each term of an expression has a common factor we divide each term by this and take it out as shown below:

Example: Factorise 9x² + 12xy

Solution:

We write 9x² + 12xy = (3x)(3x) + (3x)(4y)

Here 3x is the common to both the terms. Hence,

(3x)(3x) + (3x)(4y) = 3x(3x + 4y)

Thus, 3, x, (3x + 4y) are  the factors  of 9x² + 12xy

Method 2: Factorization[class 9] by grouping:

Sometimes, in a given expression it is not possible to take out a common factor directly. However the terms of the given expression are grouped in such a manner that we have a common factor

Example: Factorise ab + bc + ax + bx

Solution:

We have ab + bc + ax + cx = (ab + bc) +(ax +cx) = b(a + c) + x(a+ c) = (a + c)(b + x)

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Factorization[class 9] Exercise 3.2.1

1. Factorise:

(i) 5x² – 20xy

Solution:

= 5x(x – 4y)

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(ii) 2p(x + y) – 3q(x + y)

Solution:

= (x + y) (2p – 3q)

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(iii) 4(a + b)² – 6(a + b)

Solution:

= 2 (a + b) [2(a + b) – 3]

= 2 (a + b) (2a + 2b – 3)

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(iv) 18x²y – 24xyz

Solution:

= 6xy (3x – 4z)

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(v) 7xa – 70 xb

Solution:

= 7x (a – 10b)

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(vi) 13m² + 156n²

Solution:

=13(m² + 12n²)

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(vii) 3x² + 6x +6

Solution:

= 3 (x² + 2x + 2)

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(viii) 4abx² + 8abx + 120by

Solution:

= 4ab (x² + 2x +3y)

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2. Factorise

(i) x² + xy + xz + yz

Solution:

= x (x + y) + z (x+ y)

=(x + y) (x + z)

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(ii) a² – ab + ac – bx

Solution:

= a (a – b) + c (a – b)

= (a – b) (a – c)

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(iii) 4x + bx + 4b + b²

Solution:

= x (4 + b) + b (4 + b)

= (4 + b) (x + b)

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(iv) 6ax – 6a + 5a – 5

Solution:

= 6a (x – 1) + 5 (x – 1)

= (3x + 5) (x² + 1)

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(v) a³ + a²b + ab + b²

Solution:

= a² (a +b) + b (a +b)

= (a + b) (a² + b)

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(vi) 3x³ + 5x² + 3x +5

Solution:

= x² (3x +5) + 1 (3x + 5)

= (3x + 5) (x² + 1)

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(vii) y⁴ – 2y³ + y – 2

Solution:

= y³(y – 2) + 1 (y – 2)

= (y – 2) (y³ – 1)

= (y – 2) (y³ – 1³)

= (y – 2) (y + 1) (y² – y +1)

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(viii) t⁴ + t⁴ – 2t – 2

Solution:

= t³ (t + 1) – 2(t +1)

= (t – 1) (t³ – 2)

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3.2.2 Factorization of the difference of two square – Factorization[class 9]:

We know that (a + b)(a – b) = a² – b². We can use this identity to factorise many expressions involving the difference of the two squares.

Example 8: Factorise 25x² – 64y²

Solution:

We write 25x² – 64y²

25x² – 64y² = (5x)² – (8y)²

This is in the form a² – b² which factorises to (a – b)(a + b). Thus,

25x² – 64y² = (5x)² – (8y)² = (5x – 8y)(5x + 8y)

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Factorization[class 9] – Exercise 3.2.1

1. Factorize the following:

(i) 9x² – 16y²

Solution:

= (3x)² – (4y)²

Using a² – b² = (a + b) (a – b)

= (3x + 4y) (3x – 4y)

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(ii) 5x² – 7y² [ Hint : ( 𝟓 )² = 5 ]

Solution:

( 5 x)² – ( 7 y)²

Using a² – b² = (a + b) (a – b)

= ( 5 x + 7 y) ( 5 x – 7 y)

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(iii) 100 – 9x2

Solution:

= 102 – (3x)2

= (10 + 3x) (10 – 3x) [∵a² – b² = (a + b) (a – b)]

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(iv) (x + 4y)² – 4z²

Solution:

= (x + 4y)² – (2z)² [∵a² – b² = (a + b) (a – b)]

= (x + 4y + 2z) (x + 4y – 2z)

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(v) x – 64xy⁴

Solution:

= x (1 – 64y⁴)

= x (1 + (8y²)²) [∵a² – b² = (a + b) (a – b)]

= x (1 + 8y²) (1 – 8y²)

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(vi) a² + 2ab + b² – 9c²

Solution:

= (a + b)² – (3c)² [∵(a + b)² = a² + 2ab + b² ;a² – b² = (a + b) (a – b)]

= (a + b + 3c) (a + b – 3c)

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(vii) 4x² – 9y² – 2x – 3y

Solution:

= (2x)² – (3y)² – (2x + 3y)

= [(2x + 3y) (2x – 3y)] – (2x – 3y) [∵a² – b² = (a + b) (a – b)]

= (2x + 3y) (2x – 3y – 1)

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(viii) a² + b² – a + b

Solution:

= a² + b² – (a – b) [∵a² – b² = (a + b) (a – b)]

= (a + b) (a – b) – (a – b)

= (a – b) (a + b – 1)

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(ix) x⁴ – 625

Solution:

= (x²)² – 25² [∵a² – b² = (a + b) (a – b)]

= (x + 25) (x² – 25)

= (x² + 25) (x² – 5²)

= (x² + 25) (x + 5) (x – 5)

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(x) 36 (5x + y)² – 25 (4x – y)²

Solution:

= [6 (5x + y)]² – [5 (4x – y)]² [∵a² – b² = (a + b) (a – b)]

= (30x + 6y + 20x – 5y) (30x + 6y – 20x + 5y)

= (50x + y) (10x + 11y)

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(xi) (a + ^𝟏/_𝐚 )² – 4 (x – ^𝟏/_𝐱 )²

Solution:

= (a + ^𝟏/_𝐚 )² – [2 (x – ^𝟏/_𝐱 )]²

= [a+ ^𝟏/_𝐚 + 2( x− ^𝟏/_𝐱)][ a+ ^𝟏/_𝐚 – 2(x− ^𝟏/_𝐱 )]

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(xii) 12m² – 75n²

Solution:

= ( √12 m²)² – (√75 n²)² [∵a² – b² = (a + b) (a – b)]

= (2 √3m²)² – (5√3 n²)² [ 12 =2√3 and 75 = 5√3

= (2√3m² + 5√3 n²) (2√3 m² – 5√3 n²)

= √3 (2m² + 5n²) (2m – 5n) (2m + 5n)

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2. Factorize by adding and subtracting appropriate quantity.

(i) x² + 6x + 3

Solution:

= x² + ². x. 3 + 3² – 3² + 3

= (x + 3)² – 9 + 3

= (x + 3)² – 6 [∵a² – b² = (a + b) (a – b)]

= (x + 3)² – 6²

= (x + 3 + 6 ) (x + 3 – 6 )

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(ii) x² + 10x + 8

Solution:

= x² + 2. x. 5 + 5² – 5² + 8

= (x + 5)² – 25 + 8

= (x + 5)² – 17 [∵a² + 2ab + b² = (a + b)²]

= (x + 5)² – 17²

= (x + 5 + 17 ) (x + 5 – 17 )

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(iii) x² + 10x + 20

Solution:

= x² + 2. x. 5 + 5² – 5² + 20 [∵a² + 2ab + b² = (a + b)²]

= (x + 5)² – 25 + 20

= (x + 5)² – 5

= (x + 5)² – 5² [∵a² – b² = (a + b) (a – b)]

= (x + 5 + 5 ) (x + 5 – 5 )

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(iv) x² + 2x – 1

Solution:

= x² + 2. x. 1 + 1² – 1² + 1 [∵a² + 2ab + b² = (a + b)²]

= (x + 1)² – 1 + 1

= (x + 1)² – 2

= (x + 1)² – 2² [∵a² – b² = (a + b) (a – b)]

= (x + 1 + 2 ) (x + 1 – 2 )

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3. prove that (1 + ^𝐱/_𝐲 ) (1 – ^𝐱/_𝐲 ) (1 + 𝐱^𝟐/𝐲^𝟐 ) (1 + 𝐱^𝟒/𝐲^𝟒 ) + 𝐱^𝟖/𝐲^𝟖 = 1

Solution:

L.H.S = (1 + ^𝐱/_𝐲 ) (1 – ^𝐱/_𝐲 ) (1 + 𝐱^𝟐/𝐲^𝟐) (1 + 𝐱^𝟒/𝐲^𝟒 ) + 𝐱^𝟖/𝐲^𝟖

[∵a² – b² = (a + b) (a – b)]

= (1 – 𝐱^𝟐/𝐲^𝟐 ) (1 + 𝐱^𝟐/𝐲^𝟐 ) (1 + 𝐱^𝟒/𝐲^𝟒) + 𝐱^𝟖/𝐲^𝟖

= (1 – 𝐱^𝟒/𝐲^𝟒) (1 + 𝐱^𝟒/𝐲^𝟒 ) + 𝐱^𝟖/𝐲^𝟖

= (1)² – (𝐱^𝟒/𝐲^𝟒)² + 𝐱^𝟖/𝐲^𝟖

= 1 – 𝐱^𝟖/𝐲^𝟖 + 𝐱^𝟖/𝐲^𝟖

= 1 R.H.S

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4. If x = ^𝐚+𝐛/_𝐚−𝐛 and y = ^𝐚−𝐛/_𝐚+𝐛 find x² – y².

Solution:

x² – y² = (x + y) (x – y)

= (^a+b/_a−b+^a−b/_a+b)(^a+b/_a−b – ^a−b/_a+b)

= [[(a+b)²+ (a−b)² ]/(a²− b²)][[(a+b)²− (a−b)²]/(a²− b²)]

= {(a²+ 2ab + b²+ a²+ b²− 2ab)/[(a²− b²)²]} x (a² + 2ab + b² – a² – b2 + 2ab)

= [(2a²+ 2b²)4ab]/[(a²− b²)²]

= [2(a²+ b²)4ab]/(a²− b²)²

= [8ab(a²+ b²)]/(a²− b²)²

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5. If p = x – y, q = y – z and r = z – x, simplify r² – p + 2pq – q²

Solution:

(z – x)² – (x – y)² + 2 (x – y) (y – z) – (y – z)²

= z² – 2xy + x² – (x² – 2xy + y²) + 2 (xy – y² – xz + yz) – (y² + z² – 2yz)

= z² + x² – 2xz – x² + 2xy – y² + 2xy – 2y² – 2xz + 2yz – y² – z² + 2yz

= 4xy – 4xz + 4yz – 4y²

= 4[xy – xz + yz – y²]

= 4[x (y – z) – y (y – z)]

= 4 (x – y) (y – z)

= 4pq

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6. The radius of a circle is 13cm in which a chord of 10cm is drawn. Find the distance of the chord from the centre of the circle.

[Hint: use Pythagoras rule and the fact that the perpendicular line drawn from the centre of the circle to any o the chords bisects the chord]

Solution:

Let AC = 10cm is the chord of the circle with radius 13cm then using Pythagoras theory

OA² = OB² + AB²

13² = x² + 5²

x² = 13² – 5²

= (13 + 5) (13 – 5)

= 18 (8)

x² = 144 (∵ the perpendicular line drawn from the centre of the circle to a chord bisects the chord)

x = 144

x = 12 cm

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7. Is it possible to factorize x² + 4x + 20? Give reason

Solution:

x² + 4x + 20

= x² + 2. x. 2 + 2² – 2² + 20

= x² + 4x + 4 – 4 + 20

= (x + 4)² + 16

= (x + 4)² + 4²

We cannot factorize their further since this is nit of the form a² – b²

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3.2.3 More on Factorization of Trinomials – Factorization[class 9]:

Example: Factorize x² + 4√2x + 8

Solution:

We write this in the form

x² + 2(x)(2√2) + (2√2)²

This is in the form a² + 2ab + b² , where a = x, b = 2√2

But we know that a² + 2ab + b² = (a + b)²

Thus,

x² + 4√2x + 8 = (x + 2√2)²

Example: Factorise x² + 11x + 30

Solution:

Note that 11 = 5 + 6 and 30 = 5 x 6

This helps us to split the trinomial as,

x² + 11x + 30 = x² + 5x + 6x + 30

= x(x + 5)+6(x + 5)

= (x + 6)(x + 5)

Thus (x+ 5) and (x + 6) are the factors of x² + 11x + 30

Example: Factorise (x² – 2x)² – 23(x² – 2x) + 120

Solution:

We introduce a = x² – 2x . Then the expression is a² – 23a + 120. We observe that – 23 = (-15) + (-8) and 120 = (-15)(-8). Hence we can write a² – 23a + 120 = a² – 15a – 8a + 120 = a(a – 15) – 8(a – 15) = (a – 15)(a – 8)

Hence, (x² – 2x)² – 23(x² – 2x) + 120 = (x² – 2x – 15)( x² – 2x – 8)

Further we see,

(x² – 2x – 15) = x² – 5x + 3x – 15 = x(x – 5)+3(x -5) = (x + 3)(x – 5)

( x² – 2x – 8) = x² – 4x + 2x – 8 = x(x – 4)+2(x – 4) = (x – 4)(x + 2)

We hence obtain

(x² – 2x)² – 23(x² – 2x) + 120 = (x + 3)(x – 5)(x + 2)(x – 4)

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Factorization[class 9] – Exercise 3.2.3

1. Factorize

(i) x² + 9x + 18

Solution:

= x² + 6x + 3x + 18

= x (x + 6) + 3 (x + 6)

= (x + 6) (x + 3)

 

(ii) y² + 5y – 24

Solution:

= y² + 8y – 3y – 24

= y (y + 8) – 3 (y + 8)

= (y + 8) (y – 3)

 

(iii) 7y² + 49y +84

Solution:

= 7 (y² + 7y + 12)

= 7 (y² + 4y + 3y + 12)

=7 [y (y + 4) + 3 (y + 4)]

=7 (y + 4) (y + 3)

 

(iv) 40 + 3x – x²

Solution:

= – [x² – 3x – 40]

= 3 – [x² – 8x + 5x – 40]

= – [x (x – 8) + 5 (x – 8)]

= [(x – 8) (x + 5)]

= (8 – x) (x + 5)

 

(v) m² + 17mn – 84n²

Solution:

= m² + 21mn – 4mn – 84n²

= m (m + 21n) – 4n (m + 21n)

= (m + 21m) (m – 4n)

 

(vi) 117p² + 2pq – 24q²

Solution:

= 117p² + 54pq – 52pq – 24q²

= 9p (13p + 6q) – 3q (13p + 6q)

= (9p – 3q) (13p + 6q)

 

(vii) 15x² – r – 28

Solution:

= 15x² – 21x + 20x – 28

= 3x (5x – 7) + 4 (5x – 7)

= (3x + 4) (5x – 7)

 

(viii) 2x² – x – 21

Solution:

= 2x² – 7x + 6x – 21

= x (2x – 7) + 3 (2x – 7)

= (2x – 7) (x + 3)

 

(ix) 8k² – 22k – 21

Solution:

= 8k² – 28k + 6k – 21

= 4k (2k – 7) + 3(2k – 7)

= (2k – 7) (4k + 3)

 

(x) ^𝟏/_𝟑 x² – 2x – 9

Solution:

= (x²+ 6x−27)/₃

= ¹/₃ [x² – 6x – 27]

= ¹/₃ [x² – 9x + 3x – 27]

= ¹/₃ [x (x – 9) + 3 (x – 9)]

=¹/₃ (x – 9) (x + 3)

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2. Factorize

(i) √𝟓 x² + 2x – 3√𝟓

Solution:

= √5 x2 + 5x – 3x – 3√5

= √5 x (x + √5 ) – 3 ((x +√ 5 )

= (x + √5 ) (√5 x – 3)

 

(ii) √𝟑 a² + 2a – 5√𝟑

Solution:

= √3a² + 5a – 3a – 5√3 – 5√3 – √3

= a (√3 a + 5) – √3 (√3 a + 5)

= (√3 a + 5) (a – √3 )

 

(iii) 7√2 y² – 10y – 4√2

Solution:

= 7√2 y² – 14y + 4y – 4√2

= 7√2 y (y – √2) + 4 (y – 2)

= (y – √2) (7√2 y + 4)

 

(iv) 6√𝟑 z² – 47z – 5√𝟑

Solution:

= 6√3 z² – 45z – 2z – 5√3

= 3√3 z (2z –5√3) – (2z – 5√3)

= (2z – 5√3) (3√3z –1)

 

(v) 4√𝟑x² + 5x – 2√𝟑

Solution:

= 4√3 x² + 8x – 3x – 2√3

= 4x (√3x + 2) – √3(√3x + 2)

= (√3x + 2) (4x –√3)

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3. Factorize

(i) 2 (x + y)² – 9 (x + y) – 5

Solution:

Let x + y = p

= 2p² – 9p – 5

= 2p² – 10p + p – 5

= 2p(p – 5) + 1 (p – 5)

= (p – 5) (2p + 1)

= (x + y – 5) [2(x + y) + 1]

= (x + y – 5) (2x + 2y + 1)

 

(ii) 2(a – 2b)² – 25(a – 2b) + 12

Solution:

Put a – 2b = x

= 2x² – 25x + 12

= 2x² – 24x – x + 12

= 2x (x – 12) – 1 (x – 12)

= (x – 12) (2x – 1)

= (a – 2b – 12) [2(a – 2b) – 1]

= (a – 2b – 12) [2a – 4b – 1]

 

(iii) 12(z + 1)² – 25(z + 1) (x + 2) + 12 (x + 2)²

Solution:

Let z + 1 = a x + 2 = b

= 12a² – 25ab + 12b²

= 12a² – 16ab – 9ab + 12b²

= 4a (3a – 4b) – 3b (3a – 4b)

= (3a – 4b) (4a – 3b)

= [3 (z + 1) – 4 (x + 2)] [4 (z + 1) – 3 (x + 2)]

= (3z + 3 – 4x – 8) (4z + 4 3x – 2)

= (3z – 4x – 5) (4z – 3x – 2)

 

(iv) 9(2x – y) – 4(2x – y) – 13

Solution:

9(2x – y) – 4(2x – y) – 13

Put 2x – y = a

= 9a + 9a – 13a – 13

= 9a (a + 1) – 13 (a +1)

= (a + 1) (9a – 13)

= (2x – y + 1) [9 (2x – y) – 13]

= (2x – y + 1) (18x – 9y – 13)

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4. Factorize

(i) x⁴ – 3x² + 2

Solution:

Put x² = a

a² – 3a + 2

= a² – 2a – a + 2

= a (a – 2) – 1 (a – 2)

= (a – 2) (a – 1)

= (x² – 2) (x² – 1)

= (x – 2 ) (x + 2 ) (x – 1) (x + 1) [∵a² – b² = (a + b) (a – b)]

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(ii) 4x⁴ + 7x² – 2

Solution:

Put x² = a

= 4a² + 7a – 2

= 4a² + 8a – a – 2

= 4a (a + 2) – 1 (a + 2)

= (a + 2) (4a – 1) [∵a² – b² = (a + b) (a – b)]

= (x² + 2) (4x² – 1)

= (x² + 2) (2x + 1) (2x – 1)

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(iii) 3x³ – x² – 10x

Solution:

= x (3x² – x – 10)

= x (3x² – 6x + 5x – 10)

= x [3x (x – 2) + 5 (x – 2)]

= x (x – 2) (3x + 5)

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(iv) 8x³ + 2x²y – 15xy²

Solution:

= x (8x² – 2xy – 15y²)

= x (8x² – 12xy + 10xy – 15y²)

= x [4x (2x – 3y) + 5y (2x – 3y)]

= x (2x – 3y) (4x + 5y)

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(v) x⁶ – 7x³ – 8

Solution:

Put x³ = a

a³ – 7a – 8 [∵a³ + b³ = (a + b) (a² – ab + b²)]

= a³ – 8a + a – 8 [∵a³ – b³ = (a – b) (a² – ab + b²)]

= a (a – 8) + 1 (a – 8)

= (a – 8) (a + 1)

= (x³ – 3) (x³ + 1)

= (x³ – 3) (x³ + 1)

= (x + 1) (x² – x + 1) (x – 2) (x² + 2x + 4)

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3.2.4 Some Miscellaneous Factorization – Factorization[class 9]:

Consider the expression a⁴ + a²b² + b⁴ .

a⁴ + a²b² + b⁴ = (a² + b² + ab)(a² + b² – ab)

Example: Factorize 4x⁴ + 9y⁴ + 6x²y²

Solution:

a = √2x , b = √3y we obtain,

a⁴ + a²b² + b⁴ = 4x⁴ + 6x²y² + 9y⁴

We can use the factorization a⁴ + a²b² + b⁴ = (a² + b² + ab)(a² + b² – ab)

4x⁴ + 6x²y² + 9y⁴ = (2x² – √6xy + 3y²)(2x² + √6xy + 3y²)

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Factorization[class 9] – Exercise 3.2.4

1. Resolve into factors

(i) x⁴ + y⁴ – 7x² y²

Solution:

= x⁴ + y⁴ – 2x² y² – 2x² y² – 7x² y²

= x⁴ + y⁴ + 2x² y² – 9x² y²

= (x² + y²)² – (3xy)²

= (x² + y² + 3xy) (x² + y² – 3xy)

 

(ii) 4x⁴ + 25y⁴ + 10x²y²

Solution:

Take a = √2x ;b = √5y

a⁴ + b⁴ + a² b² = (a² + b² + ab) (a² + b² – ab)

4x⁴ + 25y⁴ + 10x²y² = (√2 x)⁴ + (√5 y)⁴ + 2 x² 5 y²

= [( 2 x²) + ( 5 y²) + (√10 xy] [( 2 x²) + ( 5 y²) – (√10 xy]

= (2x² +5y² + 10 xy) (2x² +5y² – 10 xy)

(iii)9a⁴+100b⁴+30a²b²

Solution:

Take a = √3a ;b = √10b

a⁴ + b⁴ + a² b² = (a² + b² + ab) (a² + b² – ab)

9a⁴+100b⁴+30a²b² = (√3a)⁴+(√10b)⁴+3a².10b²

= (3a²+10b²+30a²b²) (3a²+10b²– 30a²b²)

(iv)81a4 +9a2b2+b4

Solution:

Take a = 3a ;b = b

a⁴ + b⁴ + a² b² = (a² + b² + ab) (a² + b² – ab)

81a⁴ +9a²b²+b⁴ = (3a)⁴+(3a)²(b)²+b⁴

= (9a²+b²+3ab)(9a²+b²-3ab)

 

(v) x⁴ – 6x²y² + y⁴

= x⁴ + y⁴ + 2x²y² – 2x²y² – 6x²y²

= (x² + y²)² – ( 8 xy)²

= (x² + y² + 8 xy) (x² + y² – 8 xy)

 

1. vi) m⁴ + n⁴ – 18m²n²

= m⁴ + n⁴ + 2m²n² – 2m²n² – 18m²n²

= (m² + n²)² – 20 m²n²

= (m² + n² + √20 mn) (m² + n² – √20 mn)

 

(vii) 4m⁴ + 9n⁴ – 24m²n²

= 4m⁴ + 9n⁴ + 12 m²n² – 12 m²n² – 24m²n²

= (2m² + 3n²)² – 36mn

= (2m² + 3n² + 6mn) (2m² + 3n² – 6mn)

 

(viii) 9x⁴ + 4y⁴ + 11x²y²

= 9x⁴ + 4y⁴ + 12x²y² – 12x²y² + 11x²y²

= (3x²)² + (2y²)² + 2 (3x)² (2y)² – x²y²

= (3x² + 2y²)² – x²y²

= (3x² + 2y² + xy) (3x² + 2y² – xy)

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2. Find the factors of the following

(i) x⁴ + 9x²+81

Solution:

= x⁴+9x²+81+9x²-9x²

= (x²)²+18x²+81-9a²

=(x²+9)²-(3a)²

=( x²+9 +3a)( x²+9 – 3a)

 

(ii) a⁴ + 4a² + 16

Solution:

= a⁴ + 4a² + 16 + 4a² – 4a²

= (a²)² + 8a² + (4)² – 4a²

= (a² + 4)² – (2a)²

= (a² + 4 – 2a) (a² + 4 + 2a)

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3. Factorise the following

(i) 64a⁴ + 1

Solution:

Adding and subtracting 16a² we get

64a4 + 1 + 16a² – 16a²

= (8a²)² + 1 + 16a² – 16a²

= (8a² + 1)² – (4a)²

= (8a² + 1 + 4a) (8a² + 1 – 4a)

 

(ii) 3x⁴ + 12y⁴

Solution:

3(x⁴ + 4y⁴)

3[(x²)² + (²y²)²]

By adding and subtracting 2ab i. e.

2 x x² x 2y = 4x²y²

= 3 [(x²)² + 2(y²)² + 4x²y² – 4x²y²]

= 3 [(x² + 2y²)² – (2xy)²]

= 3 [(x² + 2y2 + 2xy) (x² + 2y² – 2xy)]

 

(iii) 4x⁴ + 81y⁴

Solution:

(2x²)² + (9y²)²

By adding and subtracting 2ab i. e.

2 x x² x 9y² = 36x²y²

= (2x²)² + (9y²)² + 36x²y² – 36x²y²

= (2x² + 9y²)² – (6xy)²

= (2x² + 9y² + 6xy) (2x² + 9y² – 6xy)

 

(iv) a⁸ – 16b⁸

Solution:

(a⁴)² – (4b⁴)²

Using a² – b² = (a + b) (a – b)

= (a⁴ + 4b⁴) (a⁴ – 4b⁴)

= (a⁴ + 4b⁴) [(a²)² – (2b²)²]

= (a⁴ + 4b⁴) (a² + 2b²) (a² – 2b²)

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3.2.5 Some More Identities – Factorization[class 9]:

Identity: a³ + b³ = (a + b)(a² – ab + b²)

We know that, (a + b)³ = a³ + b³ + 3ab(a + b)

Hence, a³ + b³ = (a + b)³ – 3ab(a + b) = (a + b)[(a+b)² – 3ab]

However,

(a + b)² – 3ab = a² + 2ab + b² – 3ab = a² – ab + b²

Using this we obtain,

a³ + b³ = (a+b)(a² – ab + b²)

Identity 2: a³ – b³ = (a – b)(a² + ab + b²)

Here we start with (a – b)³ = a³ – b³ – 3ab(a – b) . This gives, as earlier,

a³ – b³  = (a – b)³ + 3ab(a – b) = (a – b)[(a – b)² + 3ab] = (a – b)(a² + ab + b²)

Thus,

a³ – b³  = (a – b)(a² + ab + b²)

Example: Factorize x³ + 27

Solution:

We do this in several steps

Step 1: Write x³ + 27 in the form x³ + 3³ . This is in the form a³ + b³.

Step 2: We substitute a = x and b = 3 in the factorization a³ + b³ = (a + b)(a² – ab + b²)

We get,

x³ + 3³ = (x + 3)(x² – 3x + 3²)

Step 3: Simplify the expression. This gives x³ + 27 = (x + 3)(x² – 3x + 9)

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Factorization[class 9] – Exercise 3.2.5

1. Factorise:

(i) 8y³ – 1

Solution:

= (2y)³ – 1³ [∵ a³ – b³ = (a – b) (a² + ab + b²)]

= (2y – 1) [(2y)² + 2y .1 + 1²]

= (2y – 1) (4y² + 2y + 1)

 

(ii) 27x³ – 8

Solution:

= (3x)³ – 2³ [∵ a³ – b³ = (a – b) (a² + ab + b²)]

= (3x – 2) [(3x)² + 3x .2 + 2²]

= (3x – 2) (9x² + 6x + 4)

 

(iii) x³ + 8y³

Solution:

= x³ + (2y)³ [∵ a³ + b³ = (a + b) (a² – ab + b²)]

= (x + 2y) [x² – x.2y + (2y)²]

= (x + 2y) (x² – 2xy + 4y)²

 

(iv) 1 – x³

Solution:

= 1³ – x³ [∵a³ – b³ = (a – b) (a² + ab + b²)]

= (1 – x) (1² + 1.x + x²)

= (1 – x) (1 + x + x²)

 

(v) a³ b³ + c³

Solution:

= (ab)³ + c³ [∵ a³ + b³ = (a + b) (a² – ab + b²)]

= (ab + c) [(ab)² – ab .c + c²]

= (ab + c) (a²b² – abc + c²)

 

(vi) a³b – ^𝐛/_𝟔𝟒

Solution:

= b (a³– ¹/₆₄)

= b [a³ – (¹/₄ )³]

= b (a – ¹/₄) [a² + a. ¹/₄ + ( ¹/₄ )²]

= b (a – 14) (a² + ^a/₄ + ¹/₁₆ )

 

(vii) 𝐚³/_𝟖 + 1

Solution:

= ( ^a/₂ )³ + 1³ [∵ a³ + b³ = (a + b) (a² – ab + b²)]

= ( ^a/₂ + 1) [( ^a/₂ )² – ^a/₂ .1 + 1²]

= ( ^a/₂ + 1) [a²/₄− ^a/₃+ 1]

 

(viii) 3a⁶ – 𝐛^𝟔/_𝟗

Solution:

= 3(a⁶− b⁶/27)

= 3[(a²)³− (b²/₃)³]

= 3[(a²− b²/₃) (a²)³− a².b²/₃ (b²/₃)²

= 3(a²− b²/₃)( a⁴+a² . b²/₃+ b⁴/₃ )

 

(ix) 2a³ + ^𝟏/_𝟒

Solution:

= 2 (a³ + ^𝟏/_𝟒 )

= 2 (a³ + ^𝟏/₂ )³

= 2 (a + ^𝟏/₂ ) [ a² – a. ^𝟏/₂ + (^𝟏/_𝟒 )²]

= 2 (a + ^𝟏/₂ ) (a² – ^a/₂ + ^𝟏/_𝟒 )

 

(x) x³ – 512

Solution:

= x³ – 8³ [∵ a³ – b³ = (a – b) (a² + ab + b²)]

= (x – 8) (x² + 8x + 64)

 

(xi) 32x³ – 500

= 4 (8x³ – 125)

= 4 [(2x)³ – 5³]

= 4 (2x – 5) [(2x)² + 2x .5 – 5²]

= 4 (2x – 5) (4x² + 10 x + 25)

 

(xii) x⁷ + xy⁶

= x (x⁶ + y⁶)

= x [(x²)³ + (y²)³]

= x [(x² + y²) { (x²)² – x² y² + (y²)²}]

= x (x² + y²) (x⁴ – x² y² + y²)

 

(xiii) 2a⁴ – 128a

= 2a (a³ – 64)

= 2a (a³ – 4³)

= 2a (a – 4) (a² + 4a + 4²)

= 2a (a – 4) (a² + 4a + 16)

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2. Factorise

(i) (1 – a)³ + (3a)³

Solution:

Using x³ + y³ = (x + y) (x² – xy + y²)

(1 – a)³ + (3a)³

= (1 – a + 3a) [(1 – a)² – (1 – a) 3a + (3a)²]

= (1 + 2a) [(1 – a)² – 3a + 3a² + 9a²]

= (1 + 2a) (1 + a² – 2a – 3a + 12a²)

= (1 + 2a) (13a² – 5a + 1)

 

(ii) 8x³ – 27y³

Solution:

= (2x)³ – ( 3y)³

= ( 2x – 3y) [(2x)² + 2x .3y + (3y)²]

= (2x – 3y) (4x² + 6xy + 9y²)

 

(iii) z⁴ x³ + 8y³ z⁴

Solution:

= z⁴ (x³ + 8y³)

= z⁴ [x³ + (2y)³]

= z⁴ (x + 2y) [x² – x.2y + (2y)²]

= z⁴ (x + 2y) [x² – 2xy + 4y²]

 

(iv) 3(x + y)³ + ^𝟏/₉ (xy)³

Solution:

=  (x + y)³ + ¹/₂₇ (xy)³

= (x + y)³ + ( ^xy/₃ )³

= (x + y + ^xy/₃ ) (x + y)² – [ (x + y) ( ^xy/₃ )+ ( ^xy/₃)²]

= (x + y + ^xy/₃ ) (x² + y²  + 2xy – (x + y)( ^xy/₃) + x²y²/₉ )

 

(v) x⁶ + y⁶

Solution:

= (x²)³ – (y²)³

= (x² – y²) [(x²)³ + x² y²– (y²)³]

= (x² – y²) (x⁴ + x² y² + y⁴)

= (x + y) (x – y) (x² + x y + y²) (x² – xy + y²)

 

(vi) a³ – 2√𝟐 b³

Solution:

= a³ – (√2 b)³

= (a –√2b) [a² + a.√2 b + √2 b²]

= (a –√𝟐 b) (a² + √𝟐 ab + √𝟐b²)

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3. Factorize the following

(i) x⁶ – 26x³ – 27

Solution:

put x³ = a

a³ – 26a – 27

= a³ – 27a + a – 27

= a ( a – 27) + 1 (a – 27)

= (a – 27) ( a + 1)

= (x³ – 27) (x³ + 1)

= (x – 3) (x² + 3x + 9) (x – 1) (x² – x + 1)

 

(ii) z⁶ – 63z³ – 64

Solution:

put z³ = a

a³ – 63a – 64

= a³ – 64a + a – 64

= a (a – 64) + 1 (a – 64)

= (a – 64) (a + 1)

= (z³ – 64) (z³ + 1)

= (z – 4) (z² + 4z + 16) (z + 1) (z² – z + 1)

 

(iii) a³ – b³ – a + b

Solution:

= (a – b) (a² + ab + b²) – (a – b)

= (a – b) [ a² + ab + b² – 1]

 

(iv) x⁶ + 7x³ – 8

Solution:

put x³ = a

a³ + 7a – 8

= a³ + 8a – a – 8

= a (a + 8) – 1 (a + 8)

= (a – 1) (a + 8)

= (x³ – 1) (x³ + 8)

= (x – 1) (x² + x – 1) (x + 2) (x² – 2x + 4)

 

(v) a³ – ^𝟏/_𝐚^𝟑 – 2a + 𝟐𝐚

Solution:

= (a – ^𝟏/_𝐚) (a² + a. ^𝟏/_𝐚+ ^𝟏/_𝐚² ) – 2 (a – ^𝟏/_𝐚 )

= (a – ^𝟏/_𝐚 ) [a² + 1 + ^𝟏/_𝐚² – 2]

= (a – ^𝟏/_𝐚 ) (a² + ^𝟏/_𝐚^𝟐 – 1 )

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3.2.6 Factorization of some Cubic Expressions – Factorization[class 9]

1. a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)

We have,

a³ + b³ + c³ – 3abc = (a³ + b³) + c³ – 3abc

[use a³ + b³ = (a + b)³ – 3ab(a + b)]

= (a + b)³ + c³ – 3ab(a + b + c)

[use x³ + y³ = (x + y)(x² – xy + y²)]

= (a + b + c)[(a + b)² – (a + b)c + c²] – 3ab(a + b + c)

= (a + b + c)[a² + b² + 2ab – ac – bc + c² – 3ab]

=(a + b + c)(a² + b² + c² –  ab – bc – ca)

a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² –  ab – bc – ca)

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Corollary: If a + b + c = 0 then a³ + b³ + c³ = 3abc

Proof :

We use the identity

a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² –  ab – bc – ca)

Substituting a + b + c = 0, we obtain ,

a³ + b³ + c³ – 3abc = 0

Hence, a³ + b³ + c³ = 3abc

Such an identity is called a conditional identity. note that this is not valid for all choices of a, b, c it is valid only for those a, b, c satisfying a + b + c = 0

Alternatively, we can also do this as follows, Since a + b + c = 0 we get a + b = -c

Hence

(-c)³ = (a+ b)³ = a³ + b³ + 3ab(a+b) = a³ + b³ + 3ab(-c)

This simplifies to

a³ + b³ + c³ – 3abc = 0

or

a³ + b³ + c³ = 3abc

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Factorization[class 9] – Exercise 3.2.6

1. Factorise

(i) a³ – b³ – c³ – 3abc

Solution:

= a³ + (– b³)+(– c³) – 3a(–b) ( –c)

Using x³ + y³ + z³ – 3xyz = (x + y +z) (x² + y² + z² – xy – yz – xz) we get

x = a y = –b c = –c

∴ a³ + (– b³)+(– c³) – 3a(–b) (c)

= [a + (–b) +(c)] [a² + (–b)² + (–c) – a(–b) (–b)(–c) – (–c)a]

= (a – b – c) (a² + b² + c² + ab – bc + ca)

 

(ii) a³ – b³ + 8c³ + 6abc

Solution:

Using x³ + y³ + z³ – 3xyz = (x + y +z) (x² + y² + z² – xy – yz – xz) we get

x = a y = –b c = 2c

a³ + (–b)³ + (2c)³ – 3a (–b) (2c)

= [a + (– b) + 2c] [a² + (-b)² + (2c)² – a(– b) – (–b) (2c) – 2c.a]

= (a – b + 2c) (a² + b² + 4c² + ab + 2bc – ca)

 

(iii) 125a³ + b³ + 64c³ – 60abc

Solution:

Using x³ + y³ + z³ – 3xyz = (x + y +z) (x² + y² + z² – xy – yz – xz) we get

x = 5a y = b c = 4c

(5a)³ + b³ + (4c)³ – 3(5a) b (4c)

= (5a + b + 4c) [(5a)² + b² + 16c² – 5a(b) – b(4c) – 4(c) (5a)]

= (5a + b + 4c) (25a² + b²+ 16c² – 5ab – 4bc – 20ca)

 

(iv) 1 + b³ + 8c³ – 6bc

Solution:

Using x³ + y³ + z³ – 3xyz = (x + y +z) (x² + y² + z² – xy – yz – xz) we get

x = 1 y = b c = 2c

1³ + b³ + (2c)³ – 3 . 1 . b . 2c

= (1 + b + 2c) [1² + b² + (2c)² – 1 .b – b.2c – 2c.1]

= (1 + b + 2c) (1 + b² + 4c² – b – 2bc – 2c)

 

(v) 8a³ + 125b³ – 64c³ + 120abc

Solution:

Using x³ + y³ + z³ – 3xyz = (x + y +z) (x² + y² + z² – xy – yz – xz) we get

x = 2a y =5b c = –4c

(2a)³ + (5b)³ + (–4c)³ – 3(2a) (5b) (-4c)

= [2a + 5b + (–4c)] [(2a)² + (5b)² + (4c)² – (2a)(5b) – (5b)(-4c) –

(–4c) (2a)]

= (2a + 5b – 4c) (4a² + 25b² + 16c² – 1ab + 20bc + 8ca)

 

(vi) 2√𝟐 a³ + 16√𝟐 b³ + c³ – 12abc

Solution:

Using x³ + y³ + z³ – 3xyz = (x + y +z) (x² + y² + z² – xy – yz – xz) we get

x = √2a ;y =2√2 b;  c = c

(√2 a)³ + (2 √2 b)³ + c³ – 3(√2 a) (2√2 b)c

= (√2 a + 2√2 b + c) [(√2 a)² + (2√2 b)² + c² -√2 a. 2√2 b –

2√2 b . c – c. √2 a]

= (√𝟐 a + 2√𝟐 b + c) (2a² + 8b² + c² – 4ab – 2√𝟐 bc – √𝟐 ac)

 

(vii) (x – y)³ + (y – z)³ + (z – x)³

[hint: apply corollary]

Solution:

Using a + b + c =0 then a³ + b³ + c³ = 3abc using this

a = x – y b = y – z c = z – c

we have a + b + c = x – y + y – z + z – c = 0

∴ (x – y)³ + (y – z)³ + (z – x)³

= ³ (x – y) (y – z) (z – x)

 

(viii) p³ (q – r)³ + q³ (r – p)³ + (p – q)³

[Hint: apply corollary]

Solution:

Using identify if a + b + c = 0 then a³ + b³ + c³ = ³abc we get

a = p(q – r) b = q( r – p) c = r(p – q)

a + b + c = p(q – r) + q( r – p) + r(p – q)

= pq – pr – qr – pq + pr – qr

= 0

∴ p³ (q – r)³ + q³ (r – p)³ + (p – q)³

= 3p (q – r) q(r – p) r(p – q)

= 3pqr (q – r) (r – p) (p – q)

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2. Find the product using appropriate identity

(i) (a – b – c) (a² + b² + c² + ab – bc – ca)

Solution:

Using (x + y + z) (x² + y² + z² – xy – yz – zx) = x³ + y³ + x³

We get x = a y = –b z = –c

(a – b – c) (a² + b² + c² + ab – bc – ca)

= a³ + (-b)³ + (-c)³

= a³ – b³ – c³ – 3abc

 

(ii) (x – 2y – z) (x² + 4y² + z² + 2xy – 2yz – zx)

Solution:

Using (a + b + c) (a² + b² + c² – ab – bc ca) = a³ + b³ + c³ – 3abc

We get a = x b = –2y c = –z

[x + (–2y) + (–z)] [ x² +(–2y)² +(-z)² – x (-2y) – (-2y) (-z) + (-z)(x)]

= x³ + (–2y)³ + (-z)³ – 3x (–2y) (–8)

= x³ – 8y³ – z³ – 6xyz

 

(iii) (x + y – z) (x² + y² + z² – xy + yz + zx)

Solution:

Using (a + b + c) (a² + b² + c² – ab – bc ca) = a³ + b³ + c³ – 3abc

We get a = x b = y c = –z

[x + y + (–z)] [x² + y² + (–z)² – xy + y(–z) + (–z) x]

= x³ + y³ + (-z)³ – 3(x) (y) (-z)

= x³ + y³ – z³ + 3xyz

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3. Get the factorization

(a + b + c)³ – a³ – b³ – c³ = 3 (a+ b) (b + c) (c + a) writing the expression

(a + b + c)³ – a³ – b³ – c³ = {(a + b + c)³ – a³} – {b³ + c³}

Solution:

We have to prove {(a + b + c)³ – a³} – {b³ + c³}

= 3(a + b) (b+ c) (b + c)

L.H.S = {(a + b + c)³ – a³} – {b³ + c³}

Using identity a³ – b³ = (a – b) (a² + ab + b²)

= (a + b + c – a) {(a + b + c)² + (a + b + c)a + a²} – (b – c) (b² – bc + c²)

= (b + c) [(a² + b² + c² + 2ab + 2bc + 2ca + a² + ab + ac + a²) – (b² – bc + c²)

= (b + c) [(3a² + b² + c² + 3ab + 2bc + 3ca] – (b + c) (b² – bc + c²)

= [3a² + b² + c² + 3ab + 2bc + 3ca – b² + bc – c²]

= (b + c) (3a² + 3ab + 3bc + 3ca)

= (b + c) 3[a (a + b) + c (a + b)]

= 3 (b + c) (a + b) (a + c)

= 3 (a + b) (a + c) (b + c)

= R.H.S

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4. If x + y + 4 = 0, find the value of x³ + y³ – 12xy + 64.

Solution:

x + y = – 4

Cubing on both sides

(x + y)³ = (– 4)³

x³ + y³ + 3xy (x + y) = (– 4)³

x³ + y³ + 3xy (– 4)³ = – 64

x³ + y³ – 12xy + 64 = 0

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5. If x = 2y + 6, find the value of x – 8y – 36xy – 216.

Solution:

x – 2y = 6

Cubing on both sides

(x – 2y)³ = (6)³

= x³ – (2y)³ – 3x (2y) (x – 2y) = 216

= x³ – 8y³ – 6xy (6) – 216 = 0

= x³ – 8y³ – 36xy – 216 = 0

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6. Without actually calculating the cubes, find the values of the following:

(i) (–12)³ + 7³ + 5³

Solution:

We have – 12 + 7 + 5 = 0

By the identify if a + b + c = 0 then a³ + b³ + c³ – 3abc we get

(–12)³ + 7³ + 5³ = 3 (–12) (7) (5)

= -1260

 

(ii) (28)³ + (–³15)³ + (–13)³

Solution:

We find that 28 – 15 – 13 = 0

By the identify if a + b + c = 0 then a³ + b³ + c³ – ³abc we get

(28)³ + (–15)³ + (–13)³ = 3 x 28 x (–15) (–13)

= 16380

 

(iii) (–15)³ + 7³ + 8³

Solution:

Since – 15 + 7 + 8 = 0

Using the identify if a + b + c = 0 then a³ + b³ + c³ – 3abc we get

(–15)³ + 7³ + 8³ = 3 (–15) 7 x 8

= – 2520

 

(iv) (–10)³ + 3³ + 7³

Solution:

Since – 10 + 3 + 7 = 0

Using the identify if a + b + c = 0 then a³ + b³ + c³ – 3abc we get

(–10)³ + 3³ + 7³ = 3(-10) (3) (7)

= – 630

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7. Factorise the following using the identity

a³ + b³ + c³ – 3abc = ^𝟏/_𝟐 (a + b + c) [(a – b)² + (b – c)² + (c – a)²]

(i) a³ + 8b³ + c³ – 6abc

Solution:

= ¹/₂ (a + 2b + c) [(a – 2b)² + (2b – c)² + (c – a)²]

= ¹/₂ (a + 2b + c) (a² + 4b² – 4ab + 4b² + c² – 4bc + c² +

a² – 2ac)

= ¹/₂ (a + 2b + c) (2a² + 8b² + 2c² – 4ab – 4bc – 4ca)

 

(ii) 125a³ + b³ + 64c³ – 60abc

Solution:

= ¹/₂  (5a + b + 4c) [(5a – b)² + (b – 4c)² + (4c – 5a)²]

= ¹/₂  (5a + b + 4c) (25a² + b² – 10ab + b² + 16c² – 8bc + 16c²

+ 25a² – 40ac)

= ¹/₂ (5a + b + 4c) (50a² + 2b² + 32c² – 10ab – 8bc – 40ac)

 

(iii) 1³ + b³ + 8c³ – 6bc

Solution:

= ¹/₂ (1 + b + 2c) [(1 – b)² + (b – 2c)² + (2c – 1)²]

= ¹/₂ (1 + b +2c) (1 + b² – 2b + b² – 4bc + 4c² + 4c² + 1 – 4c)

= ¹/₂ (1 + b + 2c) (2 + 2b² + 8c² – 2b – 4bc – 4c)

 

(iv) 125 – 8x³ – 27y³ – 90xy

Solution:

= ¹/₂ (5 – 2x – 3y) [(5 – 2x)² + (–2x – 3y)² + (–3y – 5)²]

= ¹/₂ (5 – 2x – 3y) (25 + 4x² – 20x + 4x² + 9y² – 12xy + 9y² +

25 – 30y)

= ¹/₂ (5 –2x – 3y) (50 + 8x² + 18y² – 20x – 12xy – 30y)

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8. Factorise the following:

(i) (x – 2y)³ + (2y – 3z)³ + (3z – x)³

Solution:

We find that x – 2y + 2y – 3z + 3z – x = 0

Hence using identify if a + b + c = 0 then a³ + b³ + c³ – 3abc we get

(x – 2y)³ + (2y – 3z)³ + (3z – x)³

= 3(x – 2y) (2y – 3z) (3z – x)

 

(ii) [(𝐱^𝟐− 𝐲^𝟐)^𝟑+ (𝐲^𝟐− 𝐳^𝟐)^𝟑+(𝐳^𝟐− 𝐱^𝟐)^𝟑(𝐱 − 𝐲)^𝟑+ (𝐲 − 𝐳)^𝟑+(𝐳 − 𝐱)^𝟑]/[(x – y)³ (y – z)³ (z – x)³]

Solution:

We find that x² – y² + y² – z² + z² – y² = 0

and x – y + y – z + z – x = 0, using the identify if

a + b + c = 0 then a³ + b³ + c³ – ³abc we get

[(x²− y²)³+ (y²− z²)³+(z²− x²)³] /[(x − y)³+ (y − z)³+(z − x)³]

= [3(x²− y²)+ (y²− z²)+(z²− x²)³]/[(x – y) + (y – z)+(z – x)]

= [(x – y)(x + y)(y – z)(y + z)(z – x)(z + x)] /[(x – y)(y – z)(z – x)]

[∵ a² – b² = (a – b) (a + b)]

= (x + y) (y + z) (z + x)

 

(iii) (2x – 3y)³ + (4z – 2x)³ + (3y – 4z)³

Solution:

We find that 2x – 3y + 4z – 2x + 3y – 4z = 0

Using the identify if a + b + c = 0 then a³ + b³ + c³ – 3abc we get

(2x – 3y)³ + (4z – 2x)³ + (3y – 4z)³

= 3 (2x – 3y) (4z – 2x) (3y – 4z)

= 6 (2x – 3y) (4z – 2x) (3y – 4z)

 

(iv) (a – 3b)³ + (3b – c)³ + (c – a)³

Solution:

We find that a – 3b + 3b – c + c – a = 0

Using the identify if a + b + c = 0 then a³ + b³ + c³ – 3abc we get

(a – 3b)³ + (3b – c)³ + (c – a)³

= 3 (a – 3b) (3b – c) (c – a)

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9. Factorise the following:

(i) (x – y – z)³ – x³ + y³ + z³

Solution:

= [x + (–y) + (–z)]³ – x³ – (–y)³ – (–z)³

Using the identify if (a + b + c)³ – a³ – b³ – c³ = 3(a + b) (b +c) (c +a) we get

a = x; b = –y; c = – z

= 3[x + (–y)] [(–y) + (–z)] [(–z) + x]

= 3(x – y) (–y –z) (–z + x)

= – 3(x – y) (y + z) (x – z)

= 3(x – y) (y + z) (z – x)

 

(ii) (a – b – 1)³ – a³ + b³ + 1

Solution:

= [a + (–b)³ + (–1)³ – a³ – (–b)³ – (–1)³

Using the identify if (x + y + z)³ – x³ – y³ – z³ = 3(x + y) (y +z) (z +x) we get

x = a y = –b z = – 1

= 3 [a + (–b)] [(–b) + (–1)] [(–1) + a]

= 3(a – b) (–b – 1) (–1 + a)

= 3 (a – b) (1 + b) (a – 1)

 

(iii) (2x + y – z)³ – 8x³ – y³ + z³

Solution:

= [2x + y + (–z)]³ – (2x)³ – y³ – (–z)³

Using the identify if (a + b + c)³ – a³ – b³ – c³ = 3(a + b) (b +c) (c +a) we get

a = 2x b = y c = – z

= 3(2x + y) [y + (–z)] [(–z) + 2x]

= 3 (2x + y) (y – z) (-z + 2x)

= 3 (2x + y) (y – z) (2x – z)

 

(iv) ( a – ^𝟏/_𝟐 b + ^𝟐/_𝟑 c )³ – a³ – ^𝟏/_𝟖 b³ – ^𝟏/_𝟐𝟕 c³

Solution:

= (a – (^𝟏/_𝟐 b) + ^𝟐/_𝟑 c )3 – a³ – ( ¹/₈ b)3 – ^𝟏/_𝟐𝟕 c3

Using the identify if (x + y + z)3 – x3 – y3 – z3 = 3(x + y) (y +z) (z +x) we get

x = a ;y = – ^b/_𝟐; z = – ^c/_𝟑

= 3 (a – ^b/_𝟐 ) ( −b2 + ^c/₃ ) ( ^c/₃ – a)

= 3 (a – ^b/_𝟐 ) ( ^c/₃ – ^b/₂) (a + ^c/₃ )

 

(v) (x + y + z)³ – (2x – y)³ – (2y – z)³ – (2z – x)³

Solution:

Let us consider 2x – y + 2y – z + 2z – x = (x + y + z)

∴ The given problem can be written as

(2x – y + 2y – z + 2z – x)³ – (2x – y)³ – (2y – z)³ – (2z – x)³

Using the identify

(a + b + c)³ – a³ – b³ – c³ = 3(a + b) (b +c) (c +a) we get

a = 2x – y; b = 2y – z; c = 2z – x

= 3{(2x – y) + (2y – z)}{(2y – z) + (2z – x)}{(2z – x) + (2x – y)}

= 3 (2x – y + 2y – z) (2y – z + 2z – x) (2z – x + 2x – y)

= 3 (2x + y – z) (2y + z – x) (2z + x – y)

 

(vi) (x + y + z – 3)³ – (x – 1)³ – (y – 1)³ – (z – 1)³

Solution:

= [(x – 1) + (y – 1) + (z – 1)]³ – (x – 1)³ – (y – 1)³ – (z – 1)³

Using the identify

(a + b + c)³ – a³ – b³ – c³ = 3(a + b) (b +c) (c +a) we get

a = x – 1; b = y – 1; c = z – 1

= 3 {x – 1 (y – 1)} + {y – 1 + (z – 1)} {z – 1 + (x – 1)}

= 3 (x – 1 + y – 1) (y – 1+ z – 1) (z – 1 + x – 1)

= 3 (x + y – z) (y + z – 2) (z + x – 2)

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10. Find the factors of the following numbers

(i) 30³ – 12³ – 10³ – 8³

Solution:

We find that 12 + 10 + 8 = 30

Using the identify

(a + b + c)³ – a³ – b³ – c³ = 3(a + b) (b +c) (c +a) we get

a = 12; b = 10 ;c = 8

= 3 (12 + 10) (10 + 8) (8 + 12)

= 3 x 22 x 18 x 20

= 2⁴ x 3³. 5.11

 

(ii) 85³ – 68³ + 5³ – 22³

Solution:

We find that 68 – 5 + 22 = 85

Using the identify

(a + b + c)³ – a³ – b³ – c³ = 3(a + b) (b +c) (c +a) we get

a = 68 b = –5 c = 22

= 3 {68 + (–5)} {–5 + 22} {22 + 68} = 3 x 63 x 17 x 90

= 2.3⁵.5.7.17

 

(iii) 100³ – 49³ + 10³ – 61³

Solution:

We find that 49 – 10 + 61 = 100

Using the identify

(a + b + c)³ – a³ – b³ – c³ = 3(a + b) (b +c) (c +a) we get

a = 49; b = –10; c = 61

= 3 [49 + (10)] (–10 + 61) (61 + 49)

= 3 x 39 x 51 x 110

= 2.3³.5.11.13.17

 

11. Prove that

(x + y + z)³ – (x + y – z)³ – (y + z – x)³ – (z + x – y)³ = 24xyz

[Hint: find a relation between x + y + z and x + y – z, z + x – y]

Solution:

L.H.S. (x + y + z)³ – (x + y – z)³ – (y + z – x)³ – (z + x – y)³

= (x + y + z)³ – (x + y – z)³ – [(y + z – x)³ + (z + x – y)³]

= [x + y + z – (x + y – z)] [(x + y + z)² + (x + y + z) (x + y – z)] –

(x + y – z)²] – (y + z – x + z + x – y)] [(y + z – x)² – (y + z – x)

(z + x – z) + (z + x –y)²]

= [2z {y² + z² + x² + 2yz – 2zx – 2xy – (yz + z² – zx + xy + zx – x² – y²

– yz + xy) + z + x + y + 2xy – 2xy – 2yz}]

= [2z (3x² + 3y² + z² + 6xy) – 2z (y² + z² + x² + 2yz – 2zx – 2xy – yz – z²

+ zx – xy – zx + x² + y² + yz – xy + z² + x² + y² + 2xy – 2xy – 2yz)

= (x²z + 6y²z + 2z³ + (x + y) – 2z (3x² + 3y² – 6xy)

= 6x²z + 6y²z + 12xyz – 6x²z – 6y²z – 2z³ + 12xyz

= 24xyz

= R.H.S

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