Question 71
If a + b + c = 8 and ab + bc + ca = 20, 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
64= $$a^2 + b^2 +c^2$$ + 40
$$a^2 + b^2 +c^2$$ = 24
$$a^3 + b^3 + c^3 - 3abc$$ = (a + b + c) ($$a^2 + b^2 +c^2$$ - (ab + bc +ca))=8 (24 - 20) = 32
So, the answer would be option c) 32.
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