<s>[INST] $3x^3 = 24$ quadratic equation Completing the square I know by factoring
$$x^3 - 8 = 0\\
x-2 = 0$$
that one of the solutions is 2.
but the other solutions is $1 ± i \sqrt 3$.
Can someone explain to me how to get that?
 [/INST] Hints
$$3x^{3}=24\Leftrightarrow x^{3}-8=0$$
$$\frac{x^{3}-8}{x-2}=x^{2}+2x+4$$
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Added: The second equation means that 
$$x^{3}-8=\left( x-2\right) \left( x^{2}+2x+4\right) .$$
Thus we have
$$x^{3}-8=0\Leftrightarrow \left( x-2\right) \left( x^{2}+2x+4\right) =0,$$
whose solutions are the values of $x$ that make the factor $x-2$ or the
factor $x^{2}+2x+4$ equal to $0$. The equation $x-2=0$ has the solution $x=2$. The
solutions of the equation $x^{2}+2x+4=0$ may be found by the quadratic
formula or as follows:
$$\begin{eqnarray*}
x^{2}+2x+4 &=&0\Leftrightarrow \left( x+1\right) ^{2}+3=0\Leftrightarrow
\left( x+1\right) ^{2}=-3 \\
&\Leftrightarrow &x+1=\pm \sqrt{-3}=\pm \sqrt{3}i \\
&\Leftrightarrow &x=-1\pm \sqrt{3}i.
\end{eqnarray*}$$
So $$x^{3}-8=0\Leftrightarrow x=2\text{ or }x=-1\pm \sqrt{3}i.$$
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