#### Answer

$\left\{ -1,0,3 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Express the given equation, $
x^3-2x^2=3x
,$ in factored form. Then use the Zero Product Property by equating each factor to zero. Finally, solve each of the resulting equations.
$\bf{\text{Solution Details:}}$
Using the properties of equality, the given equation is equivalent to
\begin{array}{l}\require{cancel}
x^3-2x^2-3x=0
.\end{array}
The $GCF$ of the constants of the terms $\{
1,-2,-3
\}$ is $
1
$ since it is the highest number that can divide all the given constants. The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{
x^3,x^2,x
\}$ is $
x
.$ Hence, the entire expression has $GCF=
x
.$
Factoring the $GCF=
x
,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
x(x^2-2x-3)=0
.\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 $\left\{
1,-3
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
x(x^2+x-3x-3)=0
.\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}
x[(x^2+x)-(3x+3)]=0
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x[x(x+1)-3(x+1)]=0
.\end{array}
Factoring the $GCF=
(x+1)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
x[(x+1)(x-3)]=0
\\\\
x(x+1)(x-3)=0
.\end{array}
Equating each factor zero (Zero Product Property), then
\begin{array}{l}\require{cancel}
x=0
\\\\\text{OR}\\\\
x+1=0
\\\\\text{OR}\\\\
x-3=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x=0
\\\\\text{OR}\\\\
x+1=0
\\\\
x=-1
\\\\\text{OR}\\\\
x-3=0
\\\\
x=3
.\end{array}
Hence, the solutions are $
\left\{ -1,0,3 \right\}
.$