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Question

Find the value of a^3 + b^3 + c^3 - 3abc
given that
a+b+c = 10 & a^2 + b^2 + c^2 = 83

Answer 1

Given : a + b + c = 10 and a^2 + b^2 + c^2 = 83.

First, do this multiplication:
(a + b + c)(a^2 + b^2 + c^2)
= a^3+b^3+c^3 + a^2(b + c) + b^2(a + c) + c^2(a + b)

Now multiply through by 3 (you'll see why in a moment):
3(a + b + c)(a^2 + b^2 + c^2)
= 3(a^3+b^3+c^3) + 3a^2(b + c) + 3b^(a + c) + 3c^(a + b)
Call that Equation 1.

Now do this multiplication:
(a + b + c)^3
= a^3+b^3+c^3 + 3a^2(b+c) + 3b^2(a+c) + 3c^2(a+b) + 6abc
Call that Equation 2.

Now subtract Equation 2 from Equation 1:
3(a + b + c)(a^2 + b^2 + c^2) - (a + b + c)^3
= 3(a^3+b^3+c^3) - (a^3+b^3+c^3) - 6abc

= 2(a^3+b^3+c^3) - 6abc

Dividing through by 2 and transposing sides gives:

a^3+b^3+c^3 - 3abc
= [3(a + b + c)(a^2 + b^2 + c^2) - (a + b + c)^3] / 2

= (3 * 10 * 83 - 10^3) / 2

= (2490 - 1000) / 2

= 1490 / 2

= 745

Answer 2

a=3, b=5, c= - 7
a^2 + b^2 + c^2 = 83
a + b + c = 1
a^3 + b^3 + c^3 - 3abc = 124

I don't know if I am short sighted but I don't think it works for a + b + c = 10, since you require the sum of squares to be 83.

Answer 3

Ans: 745