Answer
$3(3x-4y)^2$
Work Step by Step
Factoring the $GCF=
3
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
27x^2-72xy+48y^2
\\\\=
3(9x^2-24xy+16y^2)
.\end{array}
Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{
expression
}$
\begin{array}{l}\require{cancel}
3(9x^2-24xy+16y^2)
\end{array} has $ac=
9(16)=144
$ and $b=
-24
.$
The two numbers with a product of $c$ and a sum of $b$ are $\left\{
-12,-12
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
3(9x^2-12xy-12xy+16y^2)
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
3[(9x^2-12xy)-(12xy-16y^2)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
3[3x(3x-4y)-4y(3x-4y)]
.\end{array}
Factoring the $GCF=
(3x-4y)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
3[(3x-4y)(3x-4y)]
\\\\=
3(3x-4y)^2
.\end{array}