After studying chapter Multiplication of Polynomials you learn to find the product of a monomial and a binomial; find the product of two polynomials; find the product of binomials using some special identities; prove some of the important identities; expand the binomial terms using the identities; solve problems under identities; prove problems under identities; prove some of the important conditional identities.
3.1.1 Introduction to Multiplication of Polynomials
Definition: An algebraic expression of the form,
f(x) = a0 + a1x + a2x2 + a3x3 + … + anxn
where, a0, a1, a2, … an are real numbers and n a non-negative integer, is called a polynomial with real coefficients or (real polynomial)
Here a0, a1, a2, … an are called the coefficients of the polynomial f(x) and a0 ,a1x , a2x2 ,a3x3, … ,anxn are called the terms of f(x). If an ≠ 0, we say that n is the degree of the polynomial f(x).
3.1.2 Some Products – Multiplication of Polynomials
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Product of a monomial by a binomial:
Consider a monomial ab and a binomial 2a + 1. We multiply them as follows:
ab(2a + 1) = ab(2a) + ab = 2a2b + ab
Example: Multiply caa and b2+2bc
Solution: Using distributive property
c2a(b2 + 2bc) = c2ab2 + c2a(2bc) = c2ab2 + 2abc3
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Product of a binomial by a binomial:
Example: Find the product (x + 2) and (x + 3)
Solution:
We use our rule: multiply term by term. Thus,
(x + 2)(x + 3) = x(x + 2) + 3(x + 2) = x2 + 2x + 3x + 6 = x2 + 5x + 6
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Some special products of binomials:
Some products occur so frequently In algebra that it is advantageous to recognize them by sight and write their product by memory. This will be particularly useful when we talk about factoring.
(a) Difference of two squares:
(a+b)(a – b) = a(a – b) + b(a – b)= a2 – ab + ab – b2 = a2 – b2
(b) Square of a binomial:
(a + b)2 = (a+b)(a+b) = a(a+b)+b(a+b) = a2 + ab + ab + b2 =a2 + 2ab + b2
(a – b)2 = (a-b)(a-b) = a(a-b)-b(a-b) = a2 – ab – ab + b2 =a2 – 2ab + b2
Example : Find the product (3 + √2)(3 + √2)
Solution:
(3 + √2)(3 + √2) = 3(3 + √2)+ √2(3 + √2) = 9 +3√2 + 3√2 + 2 = 9 + 6√2 + 2 = 11 + 6√2
Multiplication of Polynomial – Exercise 3.1.2
- Evaluate the following products:
(i) ax2 ( bx + c)
= ax2 (bx) + ax2 (c)
= abx3 + acx2
(ii) ab (a+b)
= ab (a) + ab (b)
= a2b + ab2
(iii) a2b2 (ab2+a2b)
= a2b2 (ab2) + a2b2 (a2b)
= a3b4 + a4b3
(iv) b4(b6 + b8)
= b4 (b6) + b4 (b8)
= b4 (b6) + b4 (b8)
= b10 + b12
- Evaluate the following products:
(i) (x+3) (x+2)
= (x+3) x + (x+2) x
= x2 + 3x + 2x + 6
= x2 + 5x + 6
(ii) (x+5) (x–2)
= (x+5) x + (x+5) (–2)
= x2 +5x – 2x –10
= x2 + 3x –10
(iii) ( y – 4 ) ( y + 6 )
= (y – 4) y +(y – 4)6
= y2 – 4y + 6y – 24
= y2 + 2y – 24
(iv) (a–5) (a–6)
= (a–5) a + (a–6) (–6)
= a2 – 5a – 6a + 30
= a2 – 11a +30
(v) (2x+1) (2x–3)
= (2x+1)2x + (2x+1) (–3)
= 4x2 + 2x – 6x – 3x
= 4x2 – 4x –3
(vi) ( a + b ) ( c + d )
= (a + b) c + (a+b) d
= ac + bc + ad + bd
(vii) ( 2x – 3y ) ( x – y )
= (2x – 3y) x + (2x – 3y) (–y)
= 2x2 – 3xy – 2xy +3y2
= 2x2 – 5xy + 3y2
(viii) ( √𝟕𝐱 + √𝟓 ) (√𝟓𝐱 + √𝟕 )
= (√𝟕𝐱 + √𝟓 ) (√𝟓𝐱) + (√𝟕𝐱 + √𝟓 ) (√𝟕)
= √𝟑𝟓 x2 + 5x + 7x + √𝟑𝟓
= √𝟑𝟓x2 + 12x + √𝟑𝟓
(xi) (2a+3b) (2a–3b)
= (2a+3b) 2a + (2a+3b) (–3b)
= 4a2+ 6ab – 6ab – 9b2
= 4a2 – 9b2
(xii) (6xy–5) (6xy+5)
= (6xy – 5) (6xy) + (6xy – 5) 5
= 36x2y2 – 30xy + 30xy – 25
= 36x2y2 – 25
(xiii)(2/x+3) (2/x –7)
= (2/x+3)2/x +(2/x+3)(-7)
= 4/x2 + 6/x – 14/x – 21
= 4/x2 – 8/x –21
- Expand the following using appropriate identity:
(i) (a +5)2
Using (a + b)2 = a2 +2ab +b2 we get
a = a b = 5
(a +5)2 = a2 +2.a.5 +b2
= a2 +10a +25
(ii) (2a +3)2
Using (a + b)2 = a2 +2ab +b2 we get
a = 2a; b = 3
(2a +3)2 = (2a)2 +2.2a.3 +b2
= 4a2 + 12a + 9
(iii) ( x + 𝟏/𝐱 )2
Using (a + b)2 = a2 +2ab +b2 we get
a = 2a; b = 1/x
(X + 𝟏/ )2 = x2 + 2.x. 1x + (𝟏/𝐱 )2
= x2 + 2 + (𝟏/x )2
(iv) ( √(12a) + √(6b) )2
Using (a + b)2 = a2 +2ab +b2 we get
a = √(12a) and b = √(6b)
(√(12a) + √ (6b))2= (√12a)2 + 2.√12 a + √(6b) + (√(6b))2
= 12a2 +2√72 ab +6b2
= 12a2 +2 √ (36 ×2) ab +6b2
= 12a2 + 12 √𝟐ab +6b2
(v) (𝛑 + 𝟐𝟐/7 )2
Using (a + b)2 = a2 +2ab +b2 we get
a = 𝛑 and b = 22/7
(𝛑 + 22/7 )2 = 𝛑2 + 2. 𝛑 22/7 + (22/7)2
= 𝛑2 + 44π/7 + ( 22/7)2
= 𝛑2 + 𝟒𝟒𝛑/7 + 𝟒𝟖𝟒/49
(vi) (y – 3)2
Using (a – b)2 = a2 – 2ab + b2
a = y and b = –3
(y – 3)2 = y2 – 2. y.3 + 32
= y2 – 6y + 9
(vii) (3a – 2b)2
Using (a – b)2 = a2 – 2ab + b2
a = 3a and b = –2b
(3a – 2b)2 = (3a)2 – 2.3a.2b + (2b)2
= 9a2 – 12ab + 4b2
(viii) ( y – 𝟏/y )2
Using (a – b)2 = a2 – 2ab + b2
a = y and b = 𝟏/y
(y – 𝟏/y)2 = y2 – 2.y. 𝟏/y + (𝟏/y )2
= y2 – 2 + 𝟏/y²
(ix) ( √(10x) – √(5y))2
Using (a – b)2 = a2 – 2ab + b2
a = √(10x) and b = √(5y)
(√(10x) – √(5y))2
= (√(10x))2 – 2. √(10x).√(5y)+ √(5y))2
= 10x2 – 2√50 xy + 5y2
= 10x2 – 2.5√2 xy + 5y2
= 10x2 – 10√2 xy + 5y2
(x) (𝛑 – 𝟐𝟐/7 )2
Using (a + b)2 = a2 +2ab +b2 we get
a = 𝛑 and b = 𝟐𝟐/𝟕
(𝛑 – 𝟐𝟐/ )2 = 𝛑2 – 2. 𝛑 𝟐𝟐/𝟕 + (𝟐𝟐/𝟕)2
= 𝛑2 – 44/𝟕 π + (𝟐𝟐/𝟕)2
= 𝛑2 – 𝟒𝟒𝛑/7 + 𝟒𝟖𝟒/𝟒𝟗
(xi) (2x+3) (2x+5)
Using (x + a) (x + b) = x2 + x (a + b) ab we get
x = 2x, a = 3 and b = 5
(2x+3) (2x+5) = (2x)2 +2x (3 + 5) + 3.5
= 4x2 +16x +15
(xii) (3x – 3) (3x + 4)
Using (x + a) (x + b) = x2 + x (a + b) ab we get
x = 3x, a = –3 and b = 4
(3x – 3) (3x + 4) = (3x)2 + 3x [(–3)+(4)] + (-3)4
= 9x2 + 3x –12
= 9x2 + 3x –12
- Expand :
(i) (x + 3 ) (x – 3)
Using (a + b) (a – b) = a2 – b2 we get
a = x, b = 3
(x + 3) (x – 3) = x2 – 32
= x2 – 9
(ii) (3x – 5y) (3x + 5y)
Using (a + b) (a – b) = a2 – b2 we get
a = 3x, b = 5y
(3x – 5y) (3x + 5y) = (3x)2 – (5y)2
= 9x2 – 25y2
(iii) (x/3+y/2)( x/3– y/2)
Using (a + b) (a – b) = a2 – b2 we get
a = x/3, b = y/2
(x/3+ y/2)( x/3– y/2)= (x/3 )2 – (x/3)2
= 𝐱𝟐/9-𝐲2/𝟒
(iv) (x2 + y2) (x2 – y2)
Using (a + b) (a – b) = a2 – b2 we get
a = x2, b = y2
(x2 + y2) (x2 – y2) = (x2)2 – (y2)2
= x4 – y4
(v) (a2 + 4b2) (a + 2b) (a – 2b)
Using (a + b) (a – b) = a2 – b2 for 2nd and 3rd term we get
(a2 + 4b2) (a + 2b) (a – 2b) = (a2 + 4b2) [a2 – (2b)2]
= (a2 + 4b2) (a2 – 4b2)
Using the above identity once again we get
= (a2)2 – (4b2)2
= a4 – 16b4
(vi) (x – 4) (x + 4) (x – 3) (x + 4)
Using (a + b) (a – b) = a2 – b2 we get
(x – 4) (x + 4) (x – 3) (x + 4) = (x2 – 42) (x2 – 32)
= (x2 – 16) (x2 – 9)
Using (a + b) (a – b) = x2 – x (a + b) + ab
= (x2)2 – x (16+9) +16×9
= x4 – 25x2 +144
(vii) (x – a) (x + a)(𝟏/𝐱−𝟏/𝐚)( 𝟏/𝐱 +𝟏/𝐚 )
(x2 – a2) )(( /𝐱 )2– (1/a )2)
x2 x 𝟏/𝐱 2 – x2 x 𝟏/𝐚 2 – a2 x 𝟏/𝐱 2 + a2 x 𝟏/𝐚 2
1 – x2/a2 − a2/ x2 + 1
2 – 𝐱𝟐/a𝟐 − 𝐚𝟐 /x𝟐
- Simplify the following:
(i) (2x – 3y)2 + 12xy
= (2x)2 + (3y)2 – 2.2x.3y + 12xy
= 4x2 + 9y2 -12xy +12xy
= 4x2 + 9y2
(ii) (3m + 5n)2 – (2n)2
= (3m)2 + (5n)2 + 2.3m.5n – 4n2
= 9m2 + 25n2 + 30mn – 4n2
= 9m2 + 30 mn +21n2
(iii) (4a – 7b)2 – (3a)2
= (4a)2 – 2.4a.7b + (7b)2 – (3a)2
= 16a2 – 56ab + 49b2 – 9a2
= 7a2 -56ab + 49b2
(iv) (x + 𝟏/x)2 (m + 𝟏/𝐦 )2
= (x2 + 2. x. 𝟏/x + 𝟏/x2)2 – (m2 + 2. m. 𝟏/𝐦 + 𝟏/𝐦 2)2
= x2 + 2 + 𝟏/x 2 – (m2 + 2 + 𝟏/𝐦 2)
= x2 + 2 + 𝟏/x 2 – m2 + 2 – 𝟏/𝐦 2
= x2 – m2+ 𝟏/x 𝟐 – 𝟏/𝐦 𝟐 + 4
(v) (m2 + 2n2)2 – 4m2n2
= m4 +2m4.2n2 +4n4 – 4m2 n2
= m4+ 4m2n2 + 4n2 – 4m2n2
= m4 + 4n2
(vi) (3a – 2)2 – (2a -3)2
= (9a2 – 2.3a.2 + 22) – (4a2 – 2.3a.2 + 92)
= 9a2 –12a + 4 – 4a2 + 12a – 9
= 5a2 – 5
=5(a2 – 1)
3.1.3 Identities – Multiplication of Polynomials
Consider 5x + 10 = 15 and (2x + 3)2 = 4x2 + 12x + 9. Only x = 1 satisfies the first relation, whereas any value of x satisfies the second relation. We say 5x + 10 = 15 is an equation and (2x + 3)2 = 4x2 + 12x + 9 is an identity. Thus an identity is an equation which is true for all values of the variables in it.
Proof of identities:
Identity: (x+a)(x+b)(x+c) = x3 + x2(a+b+c) + x(ab+bc+ca) + abc
Let us expand LHS, (x + a)(x + b)(x + c) = [x2 + xb + ax + ab](x + c)
= x3 + x2b + ax2 + abx + x2c + xbc + axc + abc
= x3 + x2(a + b + c) + x(ab + bc + ac) + abc
This proves the identity.
Example 7: Find the product of the binomials (x+2), (x + 3) and (x + 4)
Solution:
(x+a)(x+b)(x+c) = x3 + x2(a+b+c) + x(ab+bc+ca) + abc
Here a = 2; b = 3; c = 4 then,
(x + 2)(x + 3)(x + 4) = x3 + x2(2+3+4)+x(2×3 + 3×4 + 4×2) + (2x3x4)
= x3 + x2(9) + x(6 + 12 + 8) + (24)
= x3 + 9x2 + 26x + 24
Identity: (a+b)3 = a3 + 3a2b + 3ab2 + b3
If a = b = c then, the product (x+a)(x+b)(x+c) simply reduces to (x+a)3. In this case a+ b+ c = a+a+a = 3a, ab+bc+ca = a2 + a2 + a2 = 3a2 and abc = a3 thus we obtain,
(x + a)3 = x3 + 3x2a + 3xa2 + a3
Taking x = b in this, we obtain,
(a+b)3 = a3 + 3a2b + 3ab2 + b3
This can also be written as,
(a + b)3 = a3 + 3ab(a+ b) + b3
Example: Expand (2x + 3y)3
Solution: We use the above identity: (a + b)3 = a3 + 3a2b + 3ab2 + b3 . Take a = 2x and b = 3y and we get,
(2x + 3y)3 = (2x)3 + 3(2x)2(3y) + 3(2x)(3y)2 + (3y)3
= 8x3 + 36x2y + 54xy2 + 27y3
Multiplication of Polynomials – Exercise 3.1.3
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Find the following products:
(i)(x + 4) (x + 5) (x + 2)
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
a = 4, b = 5 and c =2
(x + 4) (x + 5) (x + 2) = x3 + x2(4 + 5 + 2) + x (4.5 + 5.2 + 2.4) + 4.5.2
= x3 + 11x2 + x (20 + 10 + 8) +40
= x3 + 11x2 + 38x +40
(ii) (y + 3) (y + 2) ( y – 1)
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc we get
x = y, a = 3, b = 2 and c = -1
(y + 3) (y + 2) (y – 1) = y3 + y2 (3 + 2 – 1) + y (3.2 + 2(-1) + (-1)3 + 3.2(-1)
= y3 + 4y2 + y (6 – 2 – 3) – 6
= y3 + 4y2 + y – 6
(iii) (a + 2) (a – 3) (a + 4)
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
x = a, a = 2, b = –3 and c = 4
(a + 2) (a – 3) (a – 4) = a3 + a2 (2 – 3 + 4) + a [2(–3) + (–3)4 + 4.2) + 2 (–3) 4
= a3 + 3a2 + a (–6 – 12 + 8) – 24
= a3 + 3a2 – 10a – 24
(iv) (m – 1) (m – 2) (m – 3)
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
x = m, a = –1, b = –2 and c = –3
(m – 1) (m – 2) (m – 3) = m3 + m2 (–1 – 2 – 3) + m [(–1) (–2) + (–2) (–3) +
(– 3)(–1)] + (–1) (–2) (–3)
= m3 + m2 (–6) + m [2 + 6 + 3] – 6
= m3 – 6m2 + 11m – 6
(v) ( √𝟐 + √𝟑) (√𝟐+ √𝟓) (√𝟐+ √𝟕 )
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
x = 2, a = 3, b = 5 and c = 7
(√2 + √3) (√2+ √5) (√2+ √7 ) = (√2)3 + (√2)2 [√3 + √5+ √7]+ √2 [√3. √5 + √5. √7 + √7. √3] + √3. √5. √7
= 2√2 + 2(√3 + √5 + √7) + √2 (15+ √35 + √21) + √105
(vi) 105 x 101 x 102
We can write this as
(100 + 5) (100 + 1) (100 + 2)
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
x = 100, a = 5, b = 1 and c = 2
(100 + 5) (100 + 1) (100 + 2) = 1003 + 1002 (5 + 1 + 2) + 100 (5.1 + 1.2
+ 2.5) + 5.1.2
= 1000000 + 10000 (8) + 100(5 + 2 +10) + 10
= 1000000 + 80000 + 1700 +10
= 1081710
(vii) 95 x 98 x 103
We can write this as
(100 – 5) (100 – 2) (100 + 3)
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
x = 100, a = -5, b = -2 and c = 3
(100 – 5) (100 – 2) (100 + 3) = 1003 + 1002 (–5 – 2 +3) + 100(–5) (–2) + (–2) 3
+ 3 (-5) + (-5) (-2) 3
= 1000000 + 10000(–4) +100 (10 – 6 –16) + 30
= 1000000 – 40000 – 1100 +30
= 958930
(viii) 1.01 x 1.02 x 1.03
We can write this as
(1 + 0.01) (1 + 0.02) (1 + 0.03)
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
x = 1, a = 0.01, b = 0.02 and c = 0.03
(1 + 0.01) (1 + 0.02) (1 + 0.03) = 13 + 12 (0.01 + 0.02 + 0.03) + 1 [(0.01) (0.02) +
(0.02) (0.03) + (0.03) (0.01)] + (0.01) (0.02) (0.03)
= 1 + 0.06 + (0.0002 + 0.0006 + 0.0003) + 0.000006
= 1.061106
- Find the coefficients of x2 and x in the following:
(i) (x + 4) (x + 1) (x + 2)
= x3 + x2 (4 + 1 + 2) + x (4.1 + 1.2 + 2.4) + 4.1.2
= x3 + 7x2 + 14x + 8
Coefficient of x2 is 7
Coefficient of x is 14
(ii) (x – 5) (x – 6) (x – 1)
= x3 + x2 (–5 – 6 –1) + x [(–5) (–6) + (–6) (–1) – 1(–5)] + (–5) (–6) (–1)
= x3– 12x2 + x (30 + 6 + 5) – 30
= x3– 12x2 + 41x – 30
Coefficient of x2 is –12 and x is 41
(iii) (2x + 1) (2x – 2) (2x – 5)
= (2x)3 + (2x)2 [1–2–5] + 2x [(1)(–2) + (–2)(–5) + (–5)(1)] + 1 (–2) (–5)
= 8x3 + 4x2 (–6) + 2x [–2 +10–5] + 10
= 8x3 – 24x2 + 6x + 10
Coefficient of x2 is –24 and x is 6
(iv) ( 𝐱/𝟐 + 1) ( 𝐱/𝟐 + 2) ( 𝐱/𝟐 + 3)
= (𝐱/ )3 + (𝐱/𝟐)2 [1 + 2 + 3] + 𝐱/𝟐 [1.2 + 2.3 + 3.1] + 1.2.3
= x3/8 + (x2/4 )(6) + 𝐱/𝟐 (2 + 6 + 3) + 6
= x3/8 + 3/2 x2 + 11/2 x + 6
Coefficient of x2 is 𝟑𝟐 and x is 𝟏𝟏𝟐
- The length, breadth and height of a cuboids are (x +3), (x – 2) and (x -1) respectively. Find its volume.
Solution:
Volume of a cuboids = length x breadth x height
V = (x +3) (x – 2) (x – 1)
Using the identity (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc we get
V = x3 + x2 (3 – 2 – 1) + x [3(–2) + (–2) (–1) + (–1)3] + 3(–2) (–1)
= x3 – 0x2 + X (–6 + 2 – 3) + 6
= x3 – 7x2 + 6
- The length, breadth and height of a metal box are cuboid are (x +5), (x – 2) and (x – 1) respectively. What is its volume?
Solution:
Volume of the metal box = length x breadth x height
V = (x +5) (x – 2) (x – 1)
Using the identity (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc we get
V = x3 + x2 + (–5 – 2 –1) + x [5(–2) + (–2) (–1) + (–1)5] + 5 (–2) (–1)
= x3 + 2x2 + x [–10 + 2 – 5] + 10
= x3 + 2x2 – 13x + 10
- Prove that
(a + b) (b + c) (c + a) = (a + b + c) (ab + bc + ca) – abc
[Hint: write a + b = a + b + c – c, b + c = a + b + c – a, c + a = a + b + c – d]
Solution:
x3 + x2 (a + b + c) + x (ab + bc + ca) + abc we get
- H.S. = (a + b + c)3 + (a + b + c)2 (–c – a – b) + (a + b + c) [(–c) (–a) +
(–a) (–b) + (–b) (–c)] – (–c) (–a) (–b)
= (a + b + c)3-(a + b + c)2[(a + b + c)] +(a + b + c) (ac + ab + bc) – abc
= (a + b + c)3– (a + b + c)3 + (a + b + c) (ac + ab + bc) – abc
= (a + b + c) (ac + ab + bc) – abc
= R. H. S.
- Find the cubes of the following:
(i) (2x +y)3
Solution:
Using (a + b)3 = a3 + 3a2b + 3ab2 + b3 we get
a = 2x, b = y
(2x +y)3 = (2x)3 + 3(2x)2 y + 3 (2x) y2 +y3
= 8x3 + 12 x2y + 6 xy2 +y3
(ii) (2x + 3y)3
Solution:
Using (a + b)3 = a3 + 3a2b + 3ab2 + b3 we get
a = 2x, b = 3y
(2x + 3y)3 = (2x)3 + 3(2x)2 (3y) + 3 (2x) (3y)2 + (3y)3
= 8x3 + 36 x2y + 54 xy2 + 27y3
(viii) 1013
Solution:
We write 101 as (100 + 1)3
Using identity (a + b)3 = a3 + 3a2b + 3ab2 + b3 we get
a = 100, b = 1
(100 + 1)3 = 1003 + 3. 1002 + 3. 100.12 + 13
= 1000000 + 30000 + 300 + 1
= 1030301
(viii) 2.13
Solution:
We write 2.1 (2 + 0.1)3
Using identity (a + b)3 = a3 + 3a2b + 3ab2 + b3 we get
a = 2, b = 0.1
(2 + 0.1)3 = 23 + 3 x 22(0.1) + 3 x 2 x (0.1)2 + (0.1)3
= 8 + 1.2 + 0.06 + 0.001
= 9.261
- Find the cubes of the following:
(i) (2a – 3b)3
Solution:
Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get
a = 2a, b = 3b
(2a – 3b)3 = (2a)3 – 3 (2a)2(3b) + 3 (2a)(3b)2 – (3b)3
= 8a3 – 36a2b + 54ab2 -27b³
(ii) ( x – 𝟏/𝐱 )3
Solution:
Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get
a = x, b = 𝟏/𝐱
(x – 𝟏/ )3 = x3 – 3x2 𝟏/𝐱 + 3x(𝟏/𝐱)3 – (𝟏/𝐱)3 = x3 – 3x + 3x/ x2 – 1/x3
= x3 – 3x + 3/x – 1/x3
(iii) (√3 x – 2)2
Solution:
Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get
a = √3 x, b = 2
(√3 x – 2)2 = (√3x)3 – 3 (√3x)2 .2 + 3. √3x x 22 – 23
= 3√3 x3 – 6. 3 x2 + 12√3 x – 8
=3√3 x3 – 18x2 + 12√3 x – 8
(iv) (2x – √5)3
Solution:
Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get
a = 2x, b = √5
(2x – √5)3 = (2x)3 – 3(2x)2 √5 + 3. 2x. √5)2 – (√5)3
= 8x3 – 12√5x2 + 30x – 5√5
(v) 493
Solution:
We can write 49 = 50 – 1
Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get
a = 50, b = 1
(50 – 1)3 = 503 – 3.502.1 + 3.50.12 – 13
= 125000 – 3 x 2500 + 150 – 1
= 125000 – 7500 +149
= 117649
(vi) 183
Solution:
Let us write 18 = 20 – 2
Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get
a = 20, b = 2
(20 – 2)3 = 203 – 3.202.2 + 3.20.22 – 23
= 8000 – 6×400 + 60×4 – 8
= 8000 – 2400 +240 – 8
= 5832
(vii) 953
Solution:
We write 95 = 100 – 5
Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get
a = 100, b = 5
(100 – 5)3 = 1003 – 3.1002.5 + 3.100.52 – 53
= 1000000 – 150000 + 7500 – 125
= 857375
(viii) 1083
Solution:
We write 1083 = (110 – 2)
Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get
a = 110, b = -2
(110 – 2)3 = 1103 – 3. (110)2.2 + 3.110×22 – 23
= 1331000 – 72600 + 1320 -8
= 1259712
- If x + 𝟏/𝐱 = 3, prove that x3 + 𝟏/𝐱𝟑 = 18.
Solution:
Given x + 1/x = 3
Cubing both sides we get
(x + 1/x )3 = 33
Using (a + b)3 = a3 + 3a2b + 3ab2 + b3 we get
a = x b = 1/x
(x + 1/x )3 = (x)3 + ( 1/x )3 + 3x. 1/x (x + 1/x )
27 = x3 + 1/x3 + 3 (3)
x3 + 1/x3 = 27 – 9
= x3 + 1/x3 = 18
- If p + q = 5 and pq = 6, find p3 + q3
Solution:
(p + q)3 = p3 + 3pq (p + q) + q3
53 = p3 + 3.6 (5) + q3
125 = p3 + 90 + q3
p3 + q3 = 125 – 90
p3 + q3 = 35
- If a – b = 3 and ab = 10, find a3 – b3
Solution:
Given a – b = 3 and ab = 10
(a – b)3 = a3 – b3 – 3ab (a – b)
33 = a3 – b3 – 3.10 (3)
27 = a3 – b3 – 90
a3 – b3 = 27 + 90
a3 – b3 = 117
- If a2 + 𝟏/𝐚𝟐 = 20 and a3 + 𝟏/𝐚𝟑 = 30, find a + 𝟏/𝐚
Solution:
a3 + 1/a3 = (a + 1/a) (a2 + 1/a2 – a x 1/a )
30 = (a + 1/a ) = (20 – 1)
30 = (a +1/a ) x 19
30/19 = a + 1/a
a + 1/a = 𝟑𝟎/𝟏𝟗
3.1.4 Square of a trinomial – Multiplication of Polynomials
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Example: Expand (x + 2y + 3z)2
Solution:
We take a = x; b = 2y ; c = 3z
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(x + 2y + 3z)2 = x2 + (2y)2 + (3z)2 + 2(x)(2y) + 2(2y)(3z) + 2(3z)(x)
= x2 + 4y2 + 9z2 + 4xy + 12yz + 6xz
Multiplication of Polynomials – Exercise 3.1.4
- Expand the following:
(i) (a + b + 2c)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get
a = a, b = b and c = 2c
(a + b + 2c)2 = a2 + b2 + (2c)2 + 2ab + 2b (2c) + 2(2c)a
= a2 + b2 + 4c2 + 2ab + 4bc + 4ca
(ii) (x + y + 3z)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca
a = x, b = y and c = 3z
(x + y + 3z)2 = x2 + y2 + (3z)2 + 2.x.y + 2y(3z) + 2.(3z)x
= x2 + y2 + 9z2 + 2xy + 6yz + 6zx
(iii) (p + q – 2r)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca
a = p, b = q and c = -2r
(p + q – 2r)2 = p2 + q2 + (-2r)2 + 2.p.q + 2q(-2r) +2(-2r)p
= p2 + q2 + 4r2 + 2pq – 4pr – 4pr
(iv) ( a/2+b/2+c/2 )2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca
a = a/2, b = b/2 and c = c/2
(a/2+ b/2+ c/2 )2 =( a/2 )2 + (b/2 )2 + (c/2 )2 + 2 (a/2 ) (b/2) + 2 (b/2 ) (c/2) +
2 (c/2 ) (a/2 )
= a2/4+b2/4+c2/4 + ab/2+bc/2+ca/2
(v) (x2 + y2 + z)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca
a = x2, b = y2 and c = z
(x2 + y2 + z)2 = (x2)2 + (y2)2 + (z)2 + 2x2y2 + 2y2z +2zx2
= x4 + y4 + z2 + 2x2y2 + 2y2z +2zx2
(vi) (m – 3 – 1/m )2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca
a = m, b = -3 and c = – 1/m
(m – 3 – 1/m )2 = m2 + (-3)2 + (1/m )2 + 2.m(-3) + 2(-3)(- 1/m ) + 2 (-1/m) m
= m2 + 9 + 1/m 2 – 6m + 6m – 2
= m2 + 1/m2 + 6/m – 6m + 7
(vii) (-a + b – c)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca
a = –a b = b c = –c
(-a + b – c)2 = (–a)2 + b2 + (–c)2 + 2(–a)b + 2b(–c) +2(–c)a
= a2 + b2 + c2 – 2ab – 2bc +2ca
(viii) (x + 5 + 1/2x )2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca
a = x; b = 5; c = 1/2x
(x + 5 + 1/2x )2 = x2 + 52 + (1/2x)2 + 2.x.5 + 2.5. 1/2x + 2(1/2x)x
= x2 + 25 + 1/4x2 +10x + 5/x + 1
= x2 + 1/4x2 +10x + 5/x + 26
- Simplify the following:
(i) (a – b + c)2 – (a – b – c)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca (a – b + c)2 – (a – b – c)2
= [ a2 + (-b)2 + c2 + 2a(-b) + 2(-b)c + 2ca ] – [a + (-b)2 + (–c)2 + 2a(–b)
+ 2(–b)(–c) + 2(–c)a]
= a2 + b2 + c2 – 2ab – 2bc +2ca – [a2 + b2 + c2 – 2ab + 2bc –2ca]
= a2 + b2 + c2 – 2ab – 2bc +2ca – a2 – b2 – c2 + 2ab – 2bc +2ca
= 4ac – 4bc
= 4c (a – b)
(ii) (3x + 4y + 5)2 – (x + 5y – 4)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca
(3x + 4y + 5)2 – (x + 5y – 4)2
= [(3x)2 + (4y)2 + 52 + 2.3x.4y + 2.4y.5 + 2.5(3x)] – [x2 + (5y)2 + (-4)2 +
- X.5y + 2.5y (-4) + 2 (-4). x]
= 9x2 + 16y2 + 25 + 24xy + 40y + 30x – [x2 + 25y2 + 16 + 10xy – 40y-8x
= 9x2 + 16y2 + 25 + 24xy + 40y + 30x – x2 – 25y2 – 16 – 10xy + 40y + 8x
= 8x2 – 9y2 + 14xy + 80y + 38x + 9
(iii) (2m – n – 3p)2 + 4mn – 6np + 12pm
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get
(2m – n – 3p)2 + 4mn – 6np + 12pm
= (2m)2 + (–n)2 + (-3p)2 + 2.2m(–n) + 2(n)(–3p) + 2 (-3p)(2m) + 4mn –
6np + 12pm
= 4m2 + n2 + 9p2 – 4mn – 6np – 12pm + 4mn – 6np + 12pm
= 4m2 + n2 + 9p2
(iv) (x + 2y + 3z + r)2 + (x + 2y + 3z – r)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get
a = (x + 2y) b = 3z c = r
(x + 2y + 3z + r)2 + (x + 2y + 3z – r)2
= (x + 2y)2 + (3z)2 + r2 + 2 (x + 2y)3z + 2.3z.r + 2.r.(x + 2y) + (x + 2y)2 +
(3z)2 – (r)2 + 2 (x + 2y) 3z + 2.3z (-r) + 2 (-r)(x + 2y)
= 2(x + 2y)2 + 9z2 + r2 + 6 (x + 2y)z + 6zr + 2r (x + 2y) + 9z2 + r2 +
6 (x + 2y) z – 6zr – 2r (x + 2y)
= 2(x2 + 2.x.2y +4y2) + 18z2 + 2r2 + 12 (x + 2y)z
= 2x2 + 8xy + 8y2 + 18z2 + 2r2 +12xz + 24 yz
= 2x2 + 8y2 + 18z2 + 2r2 + 8xy +12xy + 24 yz
- If a + b + c = 12 and a2 + b2 + c2 = 50, find ab + bc + ca.
Given a + b + c = 12 squaring both sides
(a + b + c)2 = 122
a2 + b2 + c2 + 2ab + 2bc + 2ca = 144
Given a2 + b2 + c2 = 50
50 + 2ab + 2bc + 2ca = 144
2(ab + bc + ca) = 144 – 50
2(ab + bc + ca) = 94
ab + bc + ca = 94/2
ab + bc + ca = 47
- If a2 + b2 + c2 = 35 and ab + bc + ca = 23, find all possible values of a + b +c.
Given a2 + b2 + c2 = 35 and ab + bc + ca = 23,
(a + b +c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
= a2 + b2 + c2 + 2(ab + bc + ca)
= 35 + 2 (23)
= 35 + 46
(a + b + c)2 = 81
(a + b +c) = ± √81 = ±9
- Express 4x + 9y + 16z + 12xy – 24yz – 16zx as the square of a trinomial
Using and comparing the coefficient of
(a +b+c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get
4x + 9y + 16z + 12xy – 24yz – 16zx
= (2x)2 + (3y)2 + (–4z2) + 2.2x.3y.(–4z) + 2.(–4z) + 2.(–4z).2x
= (2x + 3y – 4z)2
- If x and y are real numbers and satisfy the equation
(2x + 3y – 4z)2 + (5x – y – 4)2 = 0 find x, y.
[Hint: If a, b are real numbers such that a2 + b2 = 0, then a = b = 0.]
Given if a2 + b2 = 0, then a = b = 0
(2x + 3y – 4z)2 = 0 and (5x – y – 4)2 = 0
(2x + 3y = 5)x1 i.e., 2x + 3y = 5————-(1)
(5x – y = 4)x3 i.e., 15x – 3y = 12—————-(2)
Therefore,
2x + 3y = 5
15x – 3y = 12
—————
17x = 17
—————
x = 1
substitute the value of x in (1)
2(1)+3y = 5
2+ 3y = 5
3y = 5 – 2 = 3
y = 3/3 = 1
Thus, x = 1 and y = 1
3.1.5 Conditional Identity – Multiplication of Polynomials
Example :
If x + y = 3, prove that x3 + y3 + 3ab(a + b)
Solution:
We know that (x + y)3 = x3 + y3 + 3xy(x + y) . Substitute x + y = 3 in this identity. We get,
27 = x3 + y3 + 3xy(3) = x3 + y3 + 9xy
Multiplication of Polynomials – Exercise 3.1.5
- If a + b + c = 0, prove the following:
(i) (b + c) (b – c) + a (a + 2b) = 0
Given a + b + c = 0 the a + b = -c, b + c = -a, c + a = -b we have
L.H.S = (b + c) (b – c) + a (a + 2b)
= (-a) (b – c) + a (a + b + b)
= -ab + ac + a (-c + b)
= -ab + ab – ac + ac
= 0= R.H.S
(ii) a (a2 – bc) + b (b2 – c) + c (c2 – ab) = 0
L.H.S = a (a2 – bc) + b (b2 – c) + c (c2 – ab)
= a3 – abc + b3 – abc + c3 – abc
= a3 + b3 + c3 – 3abc
We know that if a + b + c = 0 then
a3 + b3 + c3 = 3abc
Hence we have
= 3abc – 3abc
= 0 = R.H.S
(iii) a (b2 + c2) + b (c2 + a2) + c (a2 + b2) = –3abc
L.H.S = a (b2 + c2) + b (c2 + a2) + c (a2 + b2)
= ab2 + ac2+ bc2 + ba2 + ca2 +cb2
= ab2 + ba2 + b2c + bc2 + ac2 + a2c
= ab (a + b) + bc (b + c) + ac (a + c)
= ab (–c) + bc (–a) + ac (–b)
= –abc – abc – abc
= –3abc = R.H.S
[a + b + c =0, a + b = -c, b + c = -a, c + a = -b]
(iv) (ab + bc + ca)2 = a2b2 + b2c2 + c2a2
L.H.S = (ab + bc + ca)2
= (ab)2 + (bc)2 + (ca)2 +2ab.bc + 2bc. ca + 2ca.ab
= a2b2 + b2c2 + c2a2 + 2ab2c + 2bc2a + 2ca2b
= a2b2 + b2c2 + c2a2 + 2abc + (0)
= a2b2 + b2c2 + c2a2 = R.H.S
(v) a2 – bc = b2 – ca = c2 – ab = – (ab + bc + ca)
a.a – bc = a(-b-c)-bc = -ab-ac-ba = -(ab+bc+ca)——(1)
b2 – ca = b.b – ca = b(-c-a)-ca = -bc-ab-ca = -(ab+bc+ca) ———-(2)
c2 – ab = c.c – ab = c(-a-b) – ab = -ac-bc-ab = -(ab+bc+ca) ———-(3)
From equation (1) (2) and (3)
a2 – bc = b2 – ca = c2 – ab = – (ab + bc + ca)
(vi) 2a2 + bc = (a – b) (a –c)
L.H.S = 2a2 + bc
= a2 + a2 + bc
= a2 + a x a + bc
= a2 + a (- b – c) + bc
= a2 – ab – ac + bc
= a (a – b) – c (a – b)
= (a – b) (a –c) = R.H.S
(vii) (a + b) (a – b) + ca – cb = 0
We have a + b + c = 0
a + b = c
L.H.S = (a + b) (a – b) + ac – cb
= –c (a – b) + ac – cd
= – ca + bc + ac – cb
= 0 = R.H.S
(viii) a2 + b2 + c2 = -2(ab + bc + ca)
We have a + b + c = 0
Squaring we get
(a + b + c)2 = 0
a2 + b2 + c2 + 2ab + 2bc +2ca = 0
a2 + b2 + c2 = – 2ab – 2bc – 2ca
a2 + b2 + c2 = – 2(ab + bc + ca)
Hence the proof
- Suppose a, b, c are non-zero real numbers such that a + b + c = 0,
Prove the following:
(i) 𝐚𝟐/𝐛𝐜 + 𝐛𝟐/𝐜𝐚 + 𝐜𝟐/𝐚𝐛 = 3
L.H.S = a2/bc + b2/ca + c2/ab
= (a2.a+b2.b+c2.c)/abc
= (a3+b3+c3)/abc ………. (1)
We have a + b + c = 0, a + b = –c
Cubing we get
(a + b)3 = (–c)3
a3 + b3 + 3ab + (a + b) = –c3
a3 + b3 – 3ab = –c3
a3 + b3 + c3 = 3abc ………. (2)
Substituting (2) in (1)
L.H.S = 3abc/abc = 3
(ii) ( +𝐛/𝐜 + 𝐛+𝐜/a + 𝐜+𝐚/𝐛 ) ( 𝐛/𝐜+𝐚 + 𝐜/𝐚+𝐛 + 𝐚/𝐛+𝐜 )
Whenever b + c ≠ 0, c + a ≠ 0, a + b ≠ 0
We have a + b + c =0
a + b = –c
b + c = –a
c + a = –b
L.H.S = (𝐚+𝐛/𝐜 + 𝐛+𝐜/a + 𝐜+𝐚/ ) ( 𝐛/𝐜+𝐚 + 𝐜/𝐚+𝐛 + 𝐚/𝐛+𝐜 )
= ( −c/c + −a/a + −a/b ) ( b/−b + c/−c + a/−a )
= (-1-1-1) (-1-1-1)
= (-3) (-3)
= 9 = R.H.S
(iii) 𝐚𝟐/𝟐𝐚𝟐+𝐛𝐜 + 𝐛𝟐/𝟐𝐛𝟐𝐜𝐚 + 𝐜𝟐/𝟐𝐜𝟐𝐚𝐛 = 1, provided the denominators do not become 0.
L.H.S = 𝐚𝟐/𝟐𝐚𝟐+𝐛𝐜 + 𝐛𝟐/𝟐𝐛𝟐𝐜𝐚 + 𝐜𝟐/𝟐𝐜𝟐𝐚𝐛
= 𝐚𝟐/(a-b)(a-c) + 𝐛𝟐/(b-c)(b-a) + 𝐜𝟐/(c-a)(c-b)
= 𝐚𝟐/(a-b)(a-c) – 𝐛𝟐/(a-b)(a-c + 𝐜𝟐/(a-c)(b-c)
= [a2(b−c)– b2(a−c) + c2(a−b)] /(a−b)(b−c)(a−c)
= [a2b− a2c − b2a + b2c + c2a− c2b] /(ab− b2− ac +bc) (a−c)
= [a2b− a2c − b2a + b2c + c2a− c2b]/[a2b − ab2− a2c + abc − abc + b2c + ac2]
= 1 = R.H.S
- If a + b + c = 0, prove that b2 – 4ac is a square.
We have a + b + c = 0
b = – (a + c)
Squaring on both sides
b2 = [- (a + c)]2
= (a + c)2
b2 = a2 + c2 + 2ac
Subtracting 4ac on both sides
b2 – 4ac = a2 + c2 + 2ac – 4ac
= a2 – 2ac + c2
b2 – 4ac = (a – c)2
We find that b2 – 4ac is the square of (a – c)
- If a, b, c are real numbers such that a + b + c = 2s, prove the following:
(i) s (s – a) + s (s – b) + s (s – c) = s2
L.H.S. = s (s – a) + s (s – b) + s (s – c)
= s2 – as + s2 – bs + s2 – cs
= 3s2 – as – bs – cs
= 3s2 – s (a + b + c)
= 3s2 – s (2s) (a + b + c = 2s)
= 3s2 – 2s2
= s2 = R.H.S.
(ii) s2 (s – a)2 + s (s – b)2 + s (s – c)2 = a2 + b2 + c2
L.H.S. = s2 (s – a)2 + s (s – b)2 + s (s – c)2
= s2 + s2 + a2 – 2sa + s2 + b2 + s2 + c2 – 2as – 2bs – 2cs
= 4s2 + a2 + b2 + c2 – 2s – 2bs – 2cs
= 4s2 + a2 + b2 + c2 – 2s (a + b + c)
= 4s2 + a2 + b2 + c2 – 2as (2s) (a + b + c = 2s)
= 4s2 + a2 + b2 + c2 – 4s2
= a2 + b2 + c2
= R.H.S.
(iii) (s – a) (s – b) + (s – b) (s – c) + (s – c) (s – a) + s2 = ab + bc + ca
L.H.S. = (s – a) (s – b) + (s – b) (s – c) + (s – c) (s – a) + s2
= s2 – as – bs + ab + s2 – bs – cs + bc + s2 – cs – as + ac + s2
= 4s2 – 2 as – 2bs – 2cs + ab + bc +ca
= 4s2 – 2s (a + b + c) + ab + bc +ca
= 4s2 – 2s (2s) + ab + bc +ca
= 4s2 – 4s2 + ab + bc +ca (a + b + c = 2s)
= ab + bc +ca
= R.H.S.
(iv) a2 – b2 – c2 + 2bc = 4 (s – b) ( s – c)
L.H.S = a2 – b2 – c2 + 2bc
= a2 – (b2 + c2 – 2bc)
= a2 – (b – c)2
= (a + b – c) [a – (b – c)]
= (a + b – c) (a + b – c)
= (2s – c – c) (2s – b – b)
= (2s – 2c) (2s – 2b)
= 2(s – c) (2) (s – b)
= 4 (s – c) (s – b)
= R.H.S.
- If a, b, c are real numbers, a + b + c =2s and s – a ≠ 0, s – b ≠ 0, s – c ≠ 0,
Prove that 𝐚/(𝐬−𝐚) + 𝐛/(𝐬−𝐛) + 𝐜/(𝐬−𝐜) + 2 = 𝐚𝐛𝐜/(𝐬−𝐜)( 𝐬−𝐛) (𝐬−𝐜)
L.H.S = a/(s−a) + b/(s−b) + c/(s−c) + 2
=a(s−b)( s−c) + b(s−a)(s−c) + c(s−a)( s−b) + 2(s−a)(s−b)(s−c)/ (s−a)(s−b)(s−c)
= [a(s-b)(s-c)+b(s-a)(s-c) +c(s-a)(s-b)+2(s-a)(s-b)(s-c)] /(s−a )(s−b)(S−c)
=[a(s2−bs−cs+bc) +b(s2−as−cs+ac) +c(s2−as−bs+ab) +2(s2−as−bs+ab)(s−c)] /(s−a)(s−b)(S−c)
= [as2−abs−acs+abc+bs2−abc−bcs+abc) +cs2−acs−bcs+abc+2(s3−as2−bs2+abs −cs2+acs+bcs−abc] /(s−a)(s−b)(s−c)
= [s2(a+b+c) − 2abc −2acs − 2bcs − 3abc + 2s3−2as2− 2bs2 + 2abs−2cs2+ 2acs+2bcs−2abc] /(s−a)(s−b)(s−c)
= [s2(2s) + abc+2s3−2s2 (a+b+c)]/(s−a)(s−b)(S−c)
=[2s3+ abc+2s3−2s2(2s)]/(s−a) s−b (S−c)
=[4s3+ abc−4s3]/(s−a)(s−b)(S−c)
= abc/(s−a)(s−b)(S−c)
= R.H.S.
- If a + b + c = 0, prove that a2 – bc = b2 – ca = c2 – ab = (𝐚𝟐 + 𝐛𝟐+𝐜𝟐)/𝟐
We have a + b + c = 0
Squaring will get
(a + b + c)2 = 0
a2 + b2 + c2 + 2ab + 2bc + 2ca = 0
a2 + b2 + c2 + 2b (a + c) + 2ca = 0
a2 + b2 + c2 + 2b (-b) + 2ca = 0
a2 + b2 + c2 = 2b2 – 2ca (hint: a + c = -b)
a2 + b2 + c2 = 2 (b2 – ca)
a2 + b2 + c2 = b2 – ca ……… (1)
IIIly (a + b + c)2 = 0
a2 + b2 + c2 + 2ab + 2bc + 2ca = 0
a2 + b2 + c2 + 2ab + 2c (b + a) = 0
a2 + b2 + c2 + 2ab + 2c (-c) = 0 (a + c = -c)
a2 + b2 + c2 = 2 (c2 – ab)
[a2 + b2 + c2]/2 = (c2 – ab)……………(2)
Also a2 + b2 + c2 + 2ab + 2bc + 2ca = 0
a2 + b2 + c2 + 2a (b + c) + 2bc = 0
a2 + b2 + c2 + 2a (-a) + 2bc = 0
a2 + b2 + c2 = 2 (a2 – bc) ……..(3)
From (1), (2) and (3) we get,
a2 – bc = b2 – ca = c2 – ab = [a2 + b2 + c2]/2
- If 2(a2 + b2) = (a + b)2, prove that a = b.
2a2 + 2b2 = a2 + b2 + 2ab
2a2 + 2b2 – a2 – b2 – 2ab = 0
a2 + b2 – 2ab = 0
(a – b)2 = 0
a – b =0
a = b
- If x2 – 3x + 1 = 0, prove that x2 + 𝟏/𝐱𝟐 = 7.
Given x2 – 3x + 1 = 0
x2 + 1 = 3x
x + 1/x = 3 (dividing both sides by x)
Squaring both sides we get
(x + 1/x )2 = 32 = x2 + 1/x 2 + 2x. 1/x = 9
x2 + 1/x 2 + 2 = 9
x2 + 1/x 2 = 9 – 2 = 7
