How to find all solutions for $x^3=6x+6$ Could anyone help me to find how to find all solution for $x^3=6x+6$?
 A: Use Cardano's method:
Set $y=u+v$. As you have two unknowns instead of one, you can add a constraint on $u,v$, in order to simplify the equation. This equation  becomes
$$(u+v)^3-6(u+v)-6=u^3+v^3+(u+v)(3uv-6)-6=0.$$
 Adding the condition $3uv=6$, i.e. $uv=2$, you obtain the equation $u^3+v^3-6=0$, whence the system
$$\begin{cases}u^3+v^3=6\\u^3v^3=8\end{cases}\iff u^3, v^3\;\text{ are the roots of the quadratic equation}\quad t^2-6t+8=0.$$
Now $t^2-6t+8=(t-3)^2-9+8$, so $u^3, v^3=3\pm1=2,4$ and finally
$$u,v=\sqrt[3]{2},\; \sqrt[3]{4},\enspace\text{whence}\quad y=\sqrt[3]{2}+\sqrt[3]{4}.$$
A: Set $x=ay$; the equation becomes $a^3y^3-6ay=6$. We want that
$$
\frac{a^3}{6a}=\frac{4}{3}
$$
so we can take $a=2\sqrt{2}$. Then we have
$$
16\sqrt{2}y^3-12\sqrt{2}y=6
$$
that becomes
$$
4y^3-3y=\frac{3}{4}\sqrt{2}>1
$$
Now set $y=\cosh z$, so the equation becomes
$$
4\cosh^3z-3\cosh z=\frac{3}{4}\sqrt{2}
$$
that is,
$$
\cosh3z=\frac{3}{4}\sqrt{2}
$$
Now solve
$$
e^{3z}+e^{-3z}=\frac{3}{2}\sqrt{2}
$$
and you'll have $z$ and so $y$ and finally $x$.
