Question

If a+b+c=8 and ab+bc+ac=20, find the value of a^3+b^3+c^3-3abc.

Answer 1

We have $(a+b+c)^3=8^3=512=a^3+b^3+c^3+3(a^2b+b^2c+c^2a+a^2c+b^2a+c^2b)+6abc\\ (a+b+c)(ab+bc+ca)=8*20=160=a^2b+b^2c+c^2a+a^2c+b^2a+c^2b+3abc$

Subtracting 3 times the second equation from first equation,

$512-480=a^3+b^3+c^3-3abc\\ a^3+b^3+c^3-3abc=32$

Answer.