Answer
$(xy+3)(x^2y^2-3xy+9)$
Work Step by Step
The given binomial can be written as:
$=(xy)^3+3^3$
This binomial is a sum of two cubes.
RECALL:
$a^3+b^3=(a+b)(a^2-ab+b^2)$
Factor the difference of two cubes using the formula above with $a=xy$ and $b=3$ to obtain:
$=(xy+3)[(xy)^2-xy(3)+3^2]
\\=(xy+3)(x^2y^2-3xy+9)$
