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)
Factorization[class 9] Exercise 3.2.1
- Factorise:
(i) 5x2 – 20xy
Solution:
= 5x(x – 4y)
(ii) 2p(x + y) – 3q(x + y)
Solution:
= (x + y) (2p – 3q)
(iii) 4(a + b)2 – 6(a + b)
Solution:
= 2 (a + b) [2(a + b) – 3]
= 2 (a + b) (2a + 2b – 3)
(iv) 18x2y – 24xyz
Solution:
= 6xy (3x – 4z)
(v) 7xa – 70 xb
Solution:
= 7x (a – 10b)
(vi) 13m2 + 156n2
Solution:
=13(m2 + 12n2)
(vii) 3x2 + 6x +6
Solution:
= 3 (x2 + 2x + 2)
(viii) 4abx2 + 8abx + 120by
Solution:
= 4ab (x2 + 2x +3y)
2. Factorise
(i) x2 + xy + xz + yz
Solution:
= x (x + y) + z (x+ y)
=(x + y) (x + z)
(ii) a2 – ab + ac – bx
Solution:
= a (a – b) + c (a – b)
= (a – b) (a – c)
(iii) 4x + bx + 4b + b2
Solution:
= x (4 + b) + b (4 + b)
= (4 + b) (x + b)
(iv) 6ax – 6a + 5a – 5
Solution:
= 6a (x – 1) + 5 (x – 1)
= (3x + 5) (x2 + 1)
(v) a3 + a2b + ab + b2
Solution:
= a2 (a +b) + b (a +b)
= (a + b) (a2 + b)
(vi) 3x3 + 5x2 + 3x +5
Solution:
= x2 (3x +5) + 1 (3x + 5)
= (3x + 5) (x2 + 1)
(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)
(viii) t4 + t4 – 2t – 2
Solution:
= t3 (t + 1) – 2(t +1)
= (t – 1) (t3 – 2)
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)
Factorization[class 9] – Exercise 3.2.1
- Factorize the following:
(i) 9x2 – 16y2
Solution:
= (3x)2 – (4y)2
Using a2 – b2 = (a + b) (a – b)
= (3x + 4y) (3x – 4y)
(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)
(iii) 100 – 9x2
Solution:
= 102 – (3x)2
= (10 + 3x) (10 – 3x) [∵a2 – b2 = (a + b) (a – b)]
(iv) (x + 4y)2 – 4z2
Solution:
= (x + 4y)2 – (2z)2 [∵a2 – b2 = (a + b) (a – b)]
= (x + 4y + 2z) (x + 4y – 2z)
(v) x – 64xy4
Solution:
= x (1 – 64y4)
= x (1 + (8y2)2) [∵a2 – b2 = (a + b) (a – b)]
= x (1 + 8y2) (1 – 8y2)
(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)
(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)
(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)
(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)
(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)
(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 )]
(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)
- 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 )
(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 )
(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 )
(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 )
- prove that (1 + 𝐱/𝐲 ) (1 – 𝐱/𝐲 ) (1 + 𝐱𝟐/𝐲𝟐 ) (1 + 𝐱𝟒/𝐲𝟒 ) + 𝐱𝟖/𝐲𝟖 = 1
Solution:
L.H.S = (1 + 𝐱/𝐲 ) (1 – 𝐱/𝐲 ) (1 + 𝐱𝟐/𝐲𝟐) (1 + 𝐱𝟒/𝐲𝟒 ) + 𝐱𝟖/𝐲𝟖
[∵a2 – b2 = (a + b) (a – b)]
= (1 – 𝐱𝟐/𝐲𝟐 ) (1 + 𝐱𝟐/𝐲𝟐 ) (1 + 𝐱𝟒/𝐲𝟒) + 𝐱𝟖/𝐲𝟖
= (1 – 𝐱𝟒/𝐲𝟒) (1 + 𝐱𝟒/𝐲𝟒 ) + 𝐱𝟖/𝐲𝟖
= (1)2 – (𝐱𝟒/𝐲𝟒)2 + 𝐱𝟖/𝐲𝟖
= 1 – 𝐱𝟖/𝐲𝟖 + 𝐱𝟖/𝐲𝟖
= 1 R.H.S
- 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
- 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
- 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
- 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
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)
Factorization[class 9] – Exercise 3.2.3
- 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)
- 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)
- 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)
- 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)]
(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)
(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)
(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)
(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)
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)
Factorization[class 9] – Exercise 3.2.4
- 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)
- 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)
- 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)
- 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)
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)
Factorization[class 9] – Exercise 3.2.5
- 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 – 𝐛𝟔/𝟗
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)
- 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)
- 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/𝐚𝟑 – 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/𝐚𝟐 – 1 )
3.2.6 Factorization of some Cubic Expressions – Factorization[class 9]
- 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)
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
Factorization[class 9] – Exercise 3.2.6
- 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)
- 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
- 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
- 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
- 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
- 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
- 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)
- 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) [(𝐱𝟐− 𝐲𝟐)𝟑+ (𝐲𝟐− 𝐳𝟐)𝟑+(𝐳𝟐− 𝐱𝟐)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)
- 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)
- 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
- 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