Find the limit of $6^n(2-x_n)$ where $x_n=\sqrt[3]{6+\sqrt[3]{6+\dots+\sqrt[3]{6}}}$ with $n$ roots 
Let $x_n=\sqrt[3]{6+\sqrt[3]{6+\dots+\sqrt[3]{6}}}$ where the expression in the RHS has $n$ roots.
Find the following limit: $\lim \limits_{n\to \infty}6^n(2-x_n)$

My approach: I had two approaches. The first one was the following: I showed that $x_n$ is increasing with upper bound which is equal 2 then by Weierstrass theorem its convergent with limit $2$. But we cannot deduce from that the limit of $6^n(2-x_n)$ is zero because the last expression is uncertainty.
The second one was that $x_{n+1}^3=6+x_n$ then $x_{n+1}^3-8=x_n-2$. From the last equation we get: $(x_{n+1}-2)(x_{n+1}^2+2x_{n+1}+4)=x_n-2$. I tried work out with this but I have stuck.
Would be very grateful for hints or solutions.

Lets look at the sequence
$$a_n=6^n(8-x_n^3)=6^n(2-x_n)(4+2x_n+x_n^2)$$
Now $$\lim_{n\to\infty}6^n(2-x_n)(4+2x_n+x_n^2)=\lim_{n\to\infty}12\cdot 6^n(2-x_n)$$ so lets instead look at the $\lim a_n$ and whatever the limit is just multiply it by $12$.
We have that
$$\frac{a_{n+1}}{a_n}=\frac{6^{n+1}(8-x_{n+1}^2)}{6^n(8-x_n^2)}=\frac{6(2-x_n)}{(2-x_n)(x_n^2+2x_n+4)}=\frac{6}{x_n^2+2x_n+4}$$
Since $$x_n^2+2x_n+4\geq6^{2/3}+2\cdot 6^{1/3}+4\geq 10$$ since it's increasing.
From this we can see that each $a_n$ is decreasing by $\frac{6}{10}$ at least  so we have that
$$a_0(\frac{6}{10})^n\geq a_n> 0 $$ by squeeze theorem we have that $a_n\to 0$ so the original limit tends to $0$.A side note I picked $a_n$ to make calculations easier the same could be done with the original limit.
