Solve the following cubic equation :
$x^{3} + x^{2} - 16 x = 16$

x = 1, \pm 4

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x = - 1, \pm 4

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x = 0, \pm 3

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none of these

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Solution

The correct option is A $x = - 1, \pm 4$
$x^{3} + x^{2} - 16 x = 16$

$x^{3} + x^{2} - 16 x - 16 = 0$

Let $a (x) = x^{3} + x^{2} - 16 x - 16$

$a (- 1) = - 1 + 1 + 16 - 16 = 0$

Therefore, $a (x) = (x + 1) (x^{2} - 16)$

$0 = (x + 1) (x + 4) (x - 4)$

$⟹ x = - 1, - 4, 4$

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