Factorise:16p3 − 4p
We have:
$16 p^{3}-4 p=4 p\left(4 p^{2}-1\right)$
$=4 p\left\{(2 p)^{2}-(1)^{2}\right\}$
$=4 p(2 p+1)(2 p-1)$
$\therefore 16 p^{3}-4 p=4 p(2 p+1)(2 p-1)$
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