Answer
$3(9x^2+t^2)(3x+t)(3x-t)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
243x^4-3t^4
,$ factor first the $GCF.$ Then, use the factoring of the difference of $2$ squares twice.
$\bf{\text{Solution Details:}}$
Factoring the $GCF=
3
,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
3(81x^4-t^4)
.\end{array}
Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of $2$ squares, the factored form of the expression above is
\begin{array}{l}\require{cancel}
3(9x^2+t^2)(9x^2-t^2)
.\end{array}
The last factor is also a difference of $2$ squares. Using the factoring of the difference of $2$ squares again, the expression above is equivalent to
\begin{array}{l}\require{cancel}
3(9x^2+t^2)(3x+t)(3x-t)
.\end{array}
