Why should it be $\sqrt[3]{6+x}=x$? Find all the real solutions to: 

$$x^3-\sqrt[3]{6+\sqrt[3]{x+6}}=6$$

Can you confirm the following solution? I don't understand line 3. Why should it be $\sqrt[3]{6+x}=x$?
Thank you.
$$
\begin{align}
x^3-\sqrt[3]{6+\sqrt[3]{x+6}} &= 6 \\
x^3 &= 6+ \sqrt[3]{6+\sqrt[3]{x+6}} \\
x &= \sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{x+6}}} \\
\sqrt[3]{6+x} &= x \\
x^3 &= 6+x \\
x^3-2x^2+2x^2-4x+3x-6 &= 0 \\
(x-2)(x^2+2x+3) &= 0 \\
x &= 2
\end{align}
$$

Begin as in OP $\begin{align}
x^3-\sqrt[3]{6+\sqrt[3]{x+6}} &= 6 \tag{1}\\
x^3 &= 6+ \sqrt[3]{6+\sqrt[3]{x+6}} \tag{2}\\
x &= \sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{x+6}}} \tag{3}
\end{align}$
Setting $t=\sqrt[3]{6+x}$, the equation $(3)$ rewrites
$$t^3-6=\sqrt[3]{6+\sqrt[3]{6+t}},$$ equivalent to $(1).$ Hence $t=x.$
EDIT
The functions $f:x\to x$ and $g:x\to \sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{x+6}}}\;$ are

*

*continuous on $[0,\infty)$

*increasing ($f$ is increasing trivially, $g$ is composed from increasing functions)

*$f$ is linear, $g$ is concave

*$f(0)=0<g(0), f(21)=21>g(21)$
Hence there exists unique solution to $(3).$ If $t$ and $x$ both satisfy, then they are interchangeable.
