Answer
$3(y-3x)(y^2+3xy+9x^2)$
Work Step by Step
Factor out the GCF (which is $3$) to obtain:
$=3(y^3-27x^3)$
Since $27=3^3$, the expression above is equivalent to:
$=3(y^3-3^3x^3)
\\=3[y^3-(3x)^3]$
The second factor in the expression above is a difference of two cubes.
Factor using the formula $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=y$ and $b=3x$ to obtain:
$=3(y-3x)[y^2+y(3x) + (3x)^2]
\\=3(y-3x)(y^2+3xy+9x^2)$
