Answer
$(y+3)(y-3)(x-2)(x^2+2x+4)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Group the terms of the given expression, $
x^3y^2-9x^3-8y^2+72
,$ such that the factored form of the groupings will result to a factor that is common to the entire expression. Then, use factoring by grouping. This results to a factored expression that can still be factored using the difference of $2$ squares and the difference of $2$ cubes. Use the appropriate factoring method for these.
$\bf{\text{Solution Details:}}$
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(x^3y^2-9x^3)-(8y^2-72)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x^3(y^2-9)-8(y^2-9)
.\end{array}
Factoring the $GCF=
(y^2-9)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(y^2-9)(x^3-8)
.\end{array}
The expressions $
y^2
$ and $
9
$ are both perfect squares and are separated by a minus sign. Hence, $
y^2-9
,$ 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}
(y+3)(y-3)(x^3-8)
.\end{array}
The expressions $
x^3
$ and $
8
$ are both perfect cubes (the cube root is exact). Hence, $
x^3-8
$ is a $\text{
difference
}$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent\begin{array}{l}\require{cancel}
(y+3)(y-3)(x-2)(x^2+2x+4)
.\end{array}
