Why should it be $\sqrt[3]{6+x}=x$?
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 $f(x)$.
I think the step that is is missing is
$$x = \sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{\color{red}x+6}}}$$
$$=\sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{6+\color{red}x}}}$$
$$= \sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{6+\color{red}{\sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{6+\color{blue}x}}} }}}}, $$
and we can continue to obtain the infinite expression
$$x = \sqrt[3]{6 + \sqrt[3]{6+\sqrt[3]{6+\cdots }}}$$
(This is actually invalid and handwavey and not legitimate... but let's go with it.)
Then, we can do
$$x = \sqrt[3]{6 + \color{blue}{\sqrt[3]{6 + \sqrt[3]{6+\sqrt[3]{6+\cdots }}}}}=\sqrt[3]{6 + \color{blue}x}.$$
And then, we continue.
The problem is, of course, what the #@%! is $\sqrt[3]{6 + \sqrt[3]{6+\sqrt[3]{6+\cdots }}}$ supposed to mean? Is that even well defined and can we do math on it?
And the answer is, yes.
If $a_0=6$ and $a_k = \sqrt[3]{6 + a_{k-1}}$ for $k>1$, then we can prove by induction that $0 < a_{k+1} < a_k\le 6$, with equality holding if and only if $k =0$. Therefore, $\{a_k\}$ is monotonically decreasing and bounded below, so $\lim_{n\to \infty} a_n = x$ for some real $x$, which if we wanted to, we could express as $$\sqrt[3]{6 + \sqrt[3]{6+\sqrt[3]{6+\cdots }}}$$ if we took $$\sqrt[3]{6 + \sqrt[3]{6+\sqrt[3]{6+\cdots }}}:= \lim_{n\to \infty} a_n$$ as a definition.
Now, for converging limits, $$x = \lim_{n\to \infty} a_n = \lim_{n\to \infty}a_{n+1} = \lim_{n\to\infty}\left(\sqrt[3]{6+a_n}\right) = \sqrt[3]{6 + \lim_{n\to \infty}a_n} = \sqrt[3]{6 + x}.$$
So we can do it.
I guess we can also do it without resorting to the infinite.
Let $?$ be notation for $<$ or for $>$ or for $+$, where we do not know which. However, since each of $<,>,=$ are transitive and preserved via "adding to both sides" and "cubing both sides", we may do the following manipulations.
If $x \enspace?\enspace \sqrt[3]{6+x}$, then
$$6+ x \enspace?\enspace 6 +\sqrt[3]{6+x}\Rightarrow$$
$$\sqrt[3]{6 + x} \enspace?\enspace \sqrt[3]{6 +\sqrt[3]{6+x}}\Rightarrow$$
$$x\enspace?\enspace\sqrt[3]{6 + x}\enspace?\enspace\sqrt[3]{6 +\sqrt[3]{6+x}}\Rightarrow$$
$$\sqrt[3]{6+x} \enspace?\enspace \sqrt[3]{6 +\sqrt[3]{6+x}} \enspace?\enspace \sqrt[3]{6+\sqrt[3]{6 +\sqrt[3]{6+x}}}\Rightarrow$$
$$x\enspace?\enspace\sqrt[3]{6+x} \enspace?\enspace \sqrt[3]{6 +\sqrt[3]{6+x}} \enspace?\enspace \sqrt[3]{6+\sqrt[3]{6 +\sqrt[3]{6+x}}}.$$
But we showed $$x = \sqrt[3]{6+\sqrt[3]{6 +\sqrt[3]{6+x}}}.$$
So by transitivity, $$x \enspace?\enspace x.$$ But by trichotomy, $x =x$ and $x \not < x$ and $x \not > x$. So it must be that $?$ is notation for $=$, and $x = \sqrt[6]{x+6}$.
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 $t=x.$