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.

A proof by induction.
Let:
$$f_n(x) =  \sqrt[3]{6+\sqrt[3]{6+\ldots+\sqrt[3]{6+x}}},\ g_n(x) = \sqrt{2+\sqrt{2+\ldots+\sqrt{2+x}}}$$
With $n$ terms. Then for $n=1$ you can easily solve the cubic equation to show that $f_1=g_1 $ only at $x=2$ (over the reals).
Now assume our claim is true for $n$, i.e. that $f_n(x)=g_n(x)$  iff for $x=2$. Then for $n+1$, raise to the sixth power to get that:
$$(6+f_n(x))^2=(2+g_n(x))^3$$
Clearly, this equality is true for $x=2$ since $f_n(2)=g_n(2)$. Now, if our claim is false and this equality holds for some $x_0\neq 2$, then:
$$g_n(x_0) = (6+f_n(x_0))^{2/3}-2$$
But since $dg_n/df_n < 1$ for all $x>0$, by the mean value theorem we have a contradiction.
$$$$
Thus we have proven that for any number of $n$ the only (real) solution of $f_n(x)=g_n(x)$ is $x=2$.
