The cube of the number p is 16 times the number. Find p where, p $\neq$ 0 and p $\neq$ -4.

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Solution

The correct option is D 4
We are given that
$p^{3} = 16 p$
$\Rightarrow p^{3} - 16 p = 0$
$\Rightarrow p (p^{2} - 16) = 0$
$\Rightarrow E i t h e r p = 0 o r (p^{2} - 16 = 0)$
$B u t p \neq 0 (G i v e n)$
$\Rightarrow p^{2} - 16 = 0$
$\Rightarrow p^{2} = 16$
$\Rightarrow p = \pm 4$
$B u t p \neq - 4$
$∴ p = 4$ .

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