On solving $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}}$ How do we show that there is only one solution to,$$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}}$$
I guess it is only $x=2$.
Please help.

Let $\;f(x) = \sqrt{2+x}\;$ and $\;g(x) = \sqrt[3]{6+x}$, they are strictly increasing function in $x$ when $x \ge -2$.
Since $(x+2)^3 - (x+6)^2 = (x-2)(x^2 + 7x + 14)$ and $x^2 + 7x + 14 > 0$ for all $x$,
we have
$$\begin{cases} f(x) > g(x) > 2,& x > 2\\f(x) = g(x) = 2,& x = 2\\f(x) < g(x) < 2, & x <2\end{cases}$$
So for any $x > 2$, we have
$$\begin{align}
& f(x) > g(x) > 2\\
\implies & f(f(x)) > f(g(x)) > g(g(x)) > 2\\
\implies & f(f(f(x))) > f(g(g(x)) > g(g(g(x)) > 2\\
\implies & f(f(f(f(x))) > f(g(g(g(x))) > g(g(g(g(x)))) > 2\\
\implies & f(f(f(f(x))) \ne g(g(g(g(x))))
\end{align}$$
Please note that in above deductions, we are using following reasoning repeatedly.
$$\underbrace{g\circ\cdots\circ g(x)}_{k \text{ terms}} > 2 
\implies 
f(\underbrace{g\circ\cdots\circ g(x)}_{k \text{ terms}}) >
\underbrace{g\circ\cdots\circ g(x)}_{k+1 \text{ terms}} > 2.$$
Similar logic shows that $f(f(f(f(x)))) \ne g(g(g(g(x))))$ for $x < 2$. As a result,
$x = 2$ is the only solution for the equation $f(f(f(f(x)))) = g(g(g(g(x))))$.
