Answer
$(3y+1)(9y^2-3y+1)$
Work Step by Step
The given binomial can be written as:
$(3y)^3+1^3$
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 $a=3y$ and $b=1$ to obtain:
$=(3y+1)[(3y)^2-3y(1)+1^2)
\\=(3y+1)(9y^2-3y+1)$
