Answer
$3a(9a^2+1)(3a+1)(3a-1)$
Work Step by Step
Factoring the $GCF=
3a
,$ the given $\text{
expression,
}$ $
243a^5-3a
,$ is equivalent to
\begin{align*}
3a(81a^4-1)
.\end{align*}
Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the
expression above is equivalent to
\begin{align*}
&
3a[(9a^2)^2-(1)^2]
\\&=
3a[(9a^2+1)(9a^2-1)]
\\&=
3a(9a^2+1)(9a^2-1)
.\end{align*}
Since the second factor above is still a difference of $2$ squares, then it can be further factored using the same method above. That is,
\begin{align*}
&
3a(9a^2+1)[(3a)^2-(1)^2]
\\&=
3a(9a^2+1)[(3a+1)(3a-1)]
\\&=
3a(9a^2+1)(3a+1)(3a-1)
.\end{align*}
Hence, the given expression simplifies to $
3a(9a^2+1)(3a+1)(3a-1)
$.
