Question

if a+b+c =0 , then evaluate a^3+b^3+c^3

Answer 1

a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)

Using the above identity:

a3+b3+c3=(a+b+c)(a2+b2+c2−ab−bc−ca)+3abca3+b3+c3=(a+b+c)(a2+b2+c2−ab−bc−ca)+3abc

Given a+b+c=0

So, a3+b3+c3=3abca3+b3+c3=3abc

Hope it helps!