Question 57

If a + b + c = 4 and ab + bc + ca = 2, then $$a^3 + b^3 + c^3 - 3abc$$ is equal to:

Solution

$$(a + b + c) ^ 2$$ =$$a^2 + b^2 +c^2$$ + 2ab +2bc + 2ca

16= $$a^2 + b^2 +c^2$$ + 4

$$a^2 + b^2 +c^2$$ = 12

$$a^3 + b^3 + c^3 - 3abc$$ = (a + b + c) ($$a^2 + b^2 +c^2$$ - (ab + bc +ca))=4 (12 - 2) = 40

So, the answer would be option c) 40.

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