A real solution to a cubic equation What is the easiest way to find the real solution of the equation $x^3-6x^2+6x-2=0$?
I know the solution to be $x=2+2^{2/3}+2^{1/3}$ (Mathematica) but I would like to find it analytically. If possible, not by plugging the coefficients in Cardano's or similar formula.
 A: As in the standard Cardano solution, we write $x=y+2$. So the equation becomes
$y^3+6y^2+12y+8-6(y^2+4y+4)+6(y+2)-2=0$. This simplifies to 
$y^3-6y-6=0$. In the notation of the linked article we have $p=q=-6$. 
Since $\dfrac{p^2}{4}+\dfrac{q^3}{27}=1$, the Cardano Method  works exceptionally simply.
Added: Easy, but not easiest. The method of Old John is that. 
A: If $$x^3-6x^2+6x-2=0,$$
then $$2x^3-6x^2+6x-2=x^3,$$
so that $$2(x^3-3x^2+3x-1)=x^3$$
i.e. $$2(x-1)^3 = x^3.$$
Taking cube roots of both sides gives 3 possibilities, one of which is 
$$2^{1/3}(x-1) = x$$ from which 
$$x = \frac{2^{1/3}}{2^{1/3}-1}$$
and this gives the root you want.
A: $$\begin{equation*}
x^{3}-6x^{2}+6x-2=0 \tag{1}
\end{equation*}$$
Set $x=t+2.$ Then 
$$\begin{equation*}
t^{3}-6t-6=0\tag{2}
\end{equation*}$$
Set $t=u+v.$ Then
\begin{equation*}
( u+v) ^{3}-6\left( u+v\right) -6=0,
\end{equation*}
\begin{eqnarray*}
\left( u+v\right) ^{3}-6\left( u+v\right) -6 &=&( u^{3}+v^{3}-6)
+(3u^{2}v+3uv^{2}-6u-6v) \\
&=&( u^{3}+v^{3}-6) +( 3uv-6) ( u+v) 
\end{eqnarray*}
If the auxiliary variables $u,v$ satisfy the following system, $t$ satisfies $(2)$.
\begin{equation*}
\left\{ 
\begin{array}{c}
u^{3}+v^{3}=6 \\ 
3uv=6
\end{array}
\right. \Leftrightarrow \left\{ 
\begin{array}{c}
u^{3}+v^{3}=6 \\ 
u^{3}v^{3}=8.
\end{array}
\right. 
\end{equation*}
Set $U=u^{3},V=v^{3}$. Since we know the sum and the product of $U,V$, we have
\begin{equation*}
\left\{ 
\begin{array}{c}
U+V=6 \\ 
UV=8
\end{array}
\right. \Leftrightarrow \left\{ 
\begin{array}{c}
U=u^{3}=4, \\ 
V=v^{3}=2
\end{array}
\right. \;\vee \left\{ 
\begin{array}{c}
U=u^{3}=2, \\ 
V=v^{3}=4.
\end{array}
\right. 
\end{equation*}
The pair $(u,v)=(2^{\frac{2}{3}},2^{\frac{1}{3}})$ leads to one of the solutions of $(2)$, the solution 
\begin{equation*}
t=u+v=2^{\frac{2}{3}}+2^{\frac{1}{3}}.\tag{3}
\end{equation*}
The corresponding solution of $(1)$ is thus
\begin{equation*}
x=t+2=2^{\frac{2}{3}}+2^{\frac{1}{3}}+2.\tag{4}
\end{equation*}
Remark: This agrees with Old$\ $John's creative method, because $( 2^{
\frac{2}{3}}+2^{\frac{1}{3}}+2) ( 2^{\frac{1}{3}}-1) =2^{\frac{1}{3}}$.
ADDED. The other roots of $(1)$ are complex conjugates.
