Answer
$\left(\dfrac{5}{3}x^2+3y\right)\left(\dfrac{5}{3}x^2-3y\right)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
\dfrac{25}{9}x^4-9y^2
,$ use the factoring of the difference of $2$ squares.
$\bf{\text{Solution Details:}}$
The expressions $
\dfrac{25}{9}x^4
$ and $
9y^2
$ are both perfect squares and are separated by a minus sign. Hence, $
\dfrac{25}{9}x^4-9y^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}
=\left(\dfrac{5}{3}x^2\right)^2 - \left(3y\right)^2
\\\\=\left(\dfrac{5}{3}x^2+3y\right)\left(\dfrac{5}{3}x^2-3y\right)
.\end{array}
