Question

If a + b + c = 0, then a3 + b3 + c3 – 3ab = ……………………
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Answer 1

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We know,

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

Putting a+b+c=0 on RHS, we get,

a3+b3+c3−3abc=0

⟹a3+b3+c3=3abc, Hence proved.

Answer 2

Answer:

By assumption a^3+b^3+c^3=3abc so the left hand side is 0. Therefore (a+b+c)(a^2+b^2+c^2-ab-ac-bc) = 0.