Answer
$x=\left\{ 2,-1-i\sqrt{3}, -1+ i\sqrt{3} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
x^3-8=0
,$ use the factoring of the sum or difference of $2$ cubes. Then solve the first factor using the properties of equality, while solve the second factor using the Quadratic Formula.
$\bf{\text{Solution Details:}}$
The expressions $
x^3
$ and $
8
$ are both perfect cubes (the cube root is exact). Hence, $
x^3-8=0
$ 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 to
\begin{array}{l}\require{cancel}
(x)^3-(2)^3=0
\\\\
(x-2)(x^2+2x+4)=0
.\end{array}
Equating each factor to zero (Zero Product Property), then
\begin{array}{l}\require{cancel}
x-2=0
\\\text{ OR }\\
x^2+2x+4=0
.\end{array}
Using the properties of equality, the solution to the first equation is
\begin{array}{l}\require{cancel}
x-2=0
\\\\
x=2
.\end{array}
In the second equation, $
x^2+2x+4=0
,$ $a=
1
,$ $b=
2
,$ and $c=
4
.$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then
\begin{array}{l}\require{cancel}
x=\dfrac{-2\pm\sqrt{2^2-4(1)(4)}}{2(1)}
\\\\
x=\dfrac{-2\pm\sqrt{4-16}}{2}
\\\\
x=\dfrac{-2\pm\sqrt{-12}}{2}
.\end{array}
Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy}$ and that $i=\sqrt{-1},$ the expression above is equivalent to\begin{array}{l}\require{cancel}
x=\dfrac{-2\pm\sqrt{-1}\cdot\sqrt{12}}{2}
\\\\
x=\dfrac{-2\pm i\cdot\sqrt{4\cdot3}}{2}
\\\\
x=\dfrac{-2\pm i\cdot\sqrt{(2)^2\cdot3}}{2}
\\\\
x=\dfrac{-2\pm 2i\sqrt{3}}{2}
\\\\
x=\dfrac{2(-1\pm i\sqrt{3})}{2}
\\\\
x=\dfrac{\cancel2(-1\pm i\sqrt{3})}{\cancel2}
\\\\
x=-1\pm i\sqrt{3}
.\end{array}
The solutions are $
x=\left\{ 2,-1-i\sqrt{3}, -1+ i\sqrt{3} \right\}
.$
