#### Answer

$\left\{ 0,2,4 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Express the given equation, $
x^3-6x^2=-8x
,$ 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^3-6x^2+8x=0
.\end{array}
The $GCF$ of the constants of the terms $\{
1,-6,8
\}$ 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^2-6x+8)=0
.\end{array}
Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $
1(8)=8
$ and the value of $b$ is $
-6
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
-2,-4
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
x(x^2-2x-4x+8)=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^2-2x)-(4x-8)]=0
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x[x(x-2)-4(x-2)]=0
.\end{array}
Factoring the $GCF=
(x-2)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
x[(x-2)(x-4)]=0
\\\\
x(x-2)(x-4)=0
.\end{array}
Equating each factor zero (Zero Product Property), then
\begin{array}{l}\require{cancel}
x=0
\\\\\text{OR}\\\\
x-2=0
\\\\\text{OR}\\\\
x-4=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x=0
\\\\\text{OR}\\\\
x-2=0
\\\\
x=2
\\\\\text{OR}\\\\
x-4=0
\\\\
x=4
.\end{array}
Hence, the solutions are $
\left\{ 0,2,4 \right\}
.$