Answer
$x \in \{-1-\sqrt {3}i, -1+\sqrt {3}i, 2\}$ all with a multiplicty of $1$
Work Step by Step
$f(x)=x^{3}-8$
Use the difference of cubes formula:
$a^3-b^3=(a-b)(a^2+ab+b^2)$.
We have:
,
$f(x)=(x-2)(x^{2} +2x+4)$
$x-2=0\Rightarrow x=2$
To solve $(x^{2} +2x+4)=0$, we use the quadratic formula for the quadratic equation of $ax^2+bx+c=0$, $x=\frac{-b\pm\sqrt {b^2-4ac}}{2a}$.
In this case for $x^2+2x+4$, $x=\frac{-2 \pm\sqrt {2^2-4\times1\times 4}}{2\times 1}=-1\pm\sqrt {3}i$.
$x \in \{-1-\sqrt {3}i, -1+\sqrt {3}i, 2\}$ all with a multiplicity of $1$
