Identities Involving the Product of Three Binomials :
Here we are going to see some example problems of identities involving the product of three binomials.
(x + a) (x + b) (x + c)
= x3 + (a + b + c) x2 + (ab + bc + ca)x + abc
Question 1 :
Using algebraic identity, find the coefficients of x2 , x and constant term without actual expansion.
(i) (x + 5)(x + 6)(x + 7)
Solution :
(x + 5)(x + 6)(x + 7)
x = x, a = 5, b = 6 and c = 7
= x3 + (a + b + c) x2 + (ab + bc + ca)x + abc
a + b + c = 5 + 6 + 7 = 18
ab + bc + ca = 5(6) + 6(7) + 7(5)
= 30 + 42 + 35
= 107
abc = 5(6)(7) = 210
= x3 + 18x2 + 107x + 210
Coefficient of x2 = 18
Coefficient of x = 107
Constant term = 210
(ii) (2x + 3)(2x −5)(2x −6)
Solution :
(2x + 3)(2x −5)(2x −6)
x = 2x, a = 3, b = -5 and c = -6
= x3 + (a + b + c) x2 + (ab + bc + ca)x + abc
a + b + c = 3 + (-5) + (-6) = -8
ab + bc + ca = 3(-5) + (-5)(-6) + (-6)(3)
= -15 + 30 - 18
= -3
abc = 3(-5)(-6) = 90
= x3 - 8x2 - 3x + 90
= (2x)3 - 8(2x)2 - 3(2x) + 90
= 8x3 - 32x2 - 6x + 90
Coefficient of x2 = -32
Coefficient of x = -6
Constant term = 90
Question 2 :
If (x + a)(x + b)(x + c) = x3 + 14x2 + 59x + 70 , find the value of
(i) a + b + c
(ii) (1/a) + (1/b) + (1/c)
(iii) a2 + b2 + c2
(iv) (a/bc) + (b/ac) + (c/ab)
Solution :
(x + a)(x + b)(x + c) = x3 + 14x2 + 59x + 70
x3 + (a+b+c)x2 + (ab+bc+ca)x + abc = x3 + 14x2 + 59x + 70
By equating the coefficient of x2, we get
a + b + c = 14
By equating the coefficient of x, we get
ab + bc + ca = 59
By equating constants, we get
abc = 70
(ii) (1/a) + (1/b) + (1/c)
By taking L.C.M, we get
(1/a) + (1/b) + (1/c) = (bc + ac + ab)/abc
= 59/70
(iii) a2 + b2 + c2
a2 + b2 + c2 = (a+b+c)2 - 2(ab + bc +ca)
= 142 - 2(59)
= 196 - 118
= 78
(iv) (a/bc) + (b/ac) + (c/ab)
By taking L.C.M, we get
= (a2 + b2 + c2)/abc
= 78/70
= 39/35
After having gone through the stuff given above, we hope that the students would have understood, "Identities Involving the Product of Three Binomials"
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