#### Answer

$6(a^2+4b^2)(a+2b)(a-2b)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
6a^4-96b^4
,$ factor first the $GCF.$ Then use the factoring of the sum or difference of $2$ squares.
$\bf{\text{Solution Details:}}$
The $GCF$ of the terms is $GCF=
6
$ since it is the highest number that can evenly divide (no remainder) all the given terms. Factoring the $GCF,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
6(a^4-16b^4)
.\end{array}
The expressions $
a^4
$ and $
16b^4
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
a^4-16b^4
,$ is a difference of $2$ squares. 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{array}{l}\require{cancel}
6[(a^2)^2-(4b^2)^2]
\\\\=
6(a^2+4b^2)(a^2-4b^2)
.\end{array}
The expressions $
a^2
$ and $
4b^2
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
a^2-4b^2
,$ is a difference of $2$ squares. 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{array}{l}\require{cancel}
6(a^2+4b^2)[(a)^2-(2b)^2]
\\\\=
6(a^2+4b^2)(a+2b)(a-2b)
.\end{array}