Question

If a+b+c = 3 , a^2 + b^2 + c^2 = 1 , and a^3 + b^3 + c^3 = 3 , then 6abc = ?

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Answer 1

use 2 identities :\[a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)\]\[(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc\]

Answer 2

I know both the identities. Infact, I tried to solve the same using them too, but I got stuck!

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Answer 3

Double the first one so as to get 6abc then rearrange and use the second to get the answer...