Question

If a+b+c=8 and ab+ bc+ca=20, find the value for a^{​​​​​​3}+b^{​​​​​​3}+c^{​​​​3} -3abc

the options are :-

16

32

48

64

please solve this question with explanation

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Answer 1

The two algebraic identities are

1. (a+b+c)^3 = a^3+b^3+c^3 + 3(a+b)(b+c)(a+c)

2. a^3+b^3+c^3 - 3 a b c = (a+b+c)(a^2+b^2+c^2 - a b - b c - a c)

so substituting the values from equation.

a+b+c = 8

(a+b+c)^2= a^2+b^2+c^2+2(ab+bc+ca)

a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca)

so substituting this to eqn 2 gives
a^3+b^3+c^3 - 3 a b c = (a+b+c)((a+b+c)^2 - 2(ab+bc+ca) - (a b+b c+a c))
= (a+b+c) ( (a+b+c)^2 - 3(ab+bc+ca))
=8 (8^2 - 3(20))
=8 (64-60) =8 × 4 =32.

LIKE IF SATISFIED