Answer
$81t^4-72t^2+16$
Work Step by Step
RECALL:
$x^my^m=(xy)^m$
Use the rule above to obtain:
$=[(3t+2)(3t-2)]^2$
Multiply the two binomials using the formula $(a+b)(a-b)=a^2-b^2$ to obtain:
$=[(3t)^2-2^2]^2
\\=(9t^2-4)^2$
Use the formula $(a-b)^2=a^2-2ab+b^2$ with $a=9t^2$ and $b=4$ to obtain:
$=(9t^2)^2-2(9t^2)(4)+4^2
\\=81t^4-72t^2+16$
