Question
Factorise: $16 p^3-4 p$
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Answer
$16 p^3-4 p$
$=4 p\left[4 p^2-1\right]$
$=4 p(2 p)^2-(1)^2$
$=4 p(2 p+1)(2 p-1)$
$=4 p\left[4 p^2-1\right]$
$=4 p(2 p)^2-(1)^2$
$=4 p(2 p+1)(2 p-1)$
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