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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.

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    $\begingroup$ You can write your root as $\omega+\omega^2+\omega^3$ where $\omega =2^{1/3}$. Reminds me of roots of unity. $\endgroup$ – Pedro Tamaroff Aug 27 '13 at 1:42
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    $\begingroup$ Indeed, plugging in $x=y^3+y^2+y$ and verifying $\sqrt[3]{2}$ is a zero, or $y^3-2$ is a factor solves the problem. But it is not intuitive nor naturally motivated at all $\endgroup$ – chubakueno Aug 27 '13 at 1:44
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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.

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  • $\begingroup$ Yes! ${}{}{}{}{}$ $\endgroup$ – Pedro Tamaroff Aug 27 '13 at 2:09
  • $\begingroup$ Really, simple and elegant. (+1) $\endgroup$ – Ron Gordon Aug 27 '13 at 2:15
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    $\begingroup$ Hey, that's cool, I need to write that one down, factoring in 7th heaven..:) $\endgroup$ – imranfat Aug 27 '13 at 2:37
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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.

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$$\begin{equation*} x^{3}-6x^{2}+6x-2=0 \tag{1} \end{equation*}$$ 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.

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