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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