<s>[INST] 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.
$$
\begin{align}
x^3-\sqrt[3]{6+\sqrt[3]{x+6}} &= 6 \\
x^3 &= 6+ \sqrt[3]{6+\sqrt[3]{x+6}} \\
x &= \sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{x+6}}} \\
\sqrt[3]{6+x} &= x \\
x^3 &= 6+x \\
x^3-2x^2+2x^2-4x+3x-6 &= 0 \\
(x-2)(x^2+2x+3) &= 0 \\
x &= 2
\end{align}
$$
 [/INST] So the equation is as follows
$$x=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}$$
Notice that when you repetedly put back $x$ in the RHS, you get the following infinite sum.
$$x=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+{\cdots}}}}}}}$$
From here, you can replace the part from the first cube root with $x$ to get
$$x=\sqrt[3]{6+x}$$
</s>