<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] 
Goal: Establish the fact that in order to solve $$x=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}$$
we only need to solve
$$
x=\sqrt[3]{6+x}
$$

Define $f(x)$:
$$
f(x) = \sqrt[3]{x+6}$$
From line 3, we are looking to solve:
$$
x = f(f(f(x)))$$
Notice that $f(x)$ is strictly increasing. In other words, for all $a$ and $b$ in the domain of $f(x)$, if $a > b$, then $f(a) > f(b)$.
This means that the only way in which we can get solutions for $x = f(f(f(x)))$ is if $x = f(x)$. To see why, assume that $x > f(x)$:
$$
x > f(x) \Rightarrow \\
f(x) > f(f(x)) \Rightarrow \\
f(f(x)) > f(f(f(x))) \Rightarrow \\
x > f(x) > f(f(x)) > f(f(f(x)))
$$
The statement $x = f(f(f(x)))$ is now untrue. The exact same reasoning can be used for the assumption that $x < f(x)$.
Since $x < f(x)$ and $x > f(x)$ do not give us any solutions, then $x = f(x)$ is the only case we have left.
FYI, the values of a function's domain that don't change when passed into the function are called the "fixed points" of $f(x)$. In other words, all $x$ such that $x = f(x)$ are called the fixed points of $f(x)$.
</s>