Answer
$9abc(a^2b+3c)(a^2b-c)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
9a^5b^3c+18a^3b^2c^2-27abc^3
,$ factor first the $GCF.$ Then find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping.
$\bf{\text{Solution Details:}}$
The $GCF$ of the terms is $GCF=
9abc
$ since it is the highest expression that can evenly divide (no remainder) all the given terms. Factoring the $GCF,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
9abc(a^4b^2+2a^2bc-3c^2)
.\end{array}
Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $
1(-3)=-3
$ and the value of $b$ is $
2
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
3,-1
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
9abc(a^4b^2+3a^2bc-1a^2bc-3c^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}
9abc[(a^4b^2+3a^2bc)-(1a^2bc+3c^2)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
9abc[a^2b(a^2b+3c)-c(a^2b+3c)]
.\end{array}
Factoring the $GCF=
(a^2b+3c)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
9abc[(a^2b+3c)(a^2b-c)]
\\\\=
9abc(a^2b+3c)(a^2b-c)
.\end{array}
