If $a, b,$ and $c$ are the roots of $x^{3} + 3 x^{2} + 2 = 0$ , then $a^{3} + b^{3} + c^{3} =$

33

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21

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Solution

The correct option is B $- 33$
Given, $f (x) = x^{3} + 3 x^{2} + 2$
We have,
$a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - a c)$
$\Rightarrow a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) ((a + b + c)^{2} - 3 (a b + a c + b c))$
We have,
$a + b + c = \frac{- 3}{1} = - 3$
$a b + b c + a c = 0$
$a b c = \frac{- 2}{1} = - 2$
Substituting the values, we get
$a^{3} + b^{3} + c^{3} - (3) (- 2) = (- 3) ((- 3)^{2} - 3 (0))$
$a^{3} + b^{3} + c^{3} = - 6 - 27 = - 33$

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