An unusual identity Prove that $$\left(\frac{\sqrt[3]{81+33\sqrt6}+\sqrt[3]{81-33\sqrt6}}{6}\right)^m=\left(\frac{\sqrt[5]{41+29\sqrt2}+\sqrt[5]{41-29\sqrt2}}{2}\right)^n$$ for all natural integers $m,n$.
 A: The problem statement is erroneous. The second bracket should be
$$\frac{\sqrt[5]{41+29\sqrt2}+\sqrt[5]{41-29\sqrt2}}2.$$
Note that $41+29\sqrt2=(1+\sqrt2)^5$, so $\sqrt[5]{41+29\sqrt2}=1+\sqrt2$.
Similarly $\sqrt[5]{41-29\sqrt2}=1-\sqrt2$ and
$$\frac{\sqrt[5]{41+29\sqrt2}+\sqrt[5]{41-29\sqrt2}}2=1.$$
A: Hint: for the LHS let $\,a=\sqrt[3]{81+33\sqrt6}\,, \,b=\sqrt[3]{81-33\sqrt6}\,$ then $\,a^3+b^3=162\,$ and $\,ab=3\,$, so: $$\,(a+b)^3=a^3+b^3+3ab(a+b)=162+9(a+b)\,$$
But the equation $x^3-9x-162=0$ has only one real root which is $x=6$, therefore $a+b=6$.

[ EDIT ] As pointed out by @LordSharktheUnknown, the RHS would need to be $5^{th}$ roots in order for the equality to hold, rather than cube roots as posted.

An argument like the above works in that case, too. Let $a=\sqrt[5]{41+29\sqrt2}\,$, $b=\sqrt[5]{41-29\sqrt2}$ then $a^5+b^5=82$ and $ab=-1\,$, so:
$$
\begin{align}
(a+b)^5 &= a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b) \\
 &= a^5+b^5 +5ab\big((a+b)^3-3ab(a+b)\big)+10a^2b^2(a+b) \\
 &= 82 - 5\big((a+b)^3+3(a+b)\big)+10(a+b) \\
 &= 82 - 5(a+b)^3 - 5(a+b)
\end{align}
$$
But the equation $x^5+5x^3+5x-82=0$ has only one real root $x=2$, therefore $a+b=2$.
A: Hint: if the left side doesn't depend on $m$, the quantity inside parentheses must be either $0$ or $1$.  Similarly for the right side.
Further hint: $\sqrt[3]{81 \pm 33 \sqrt{6}}$ have the form $a \pm b \sqrt{6}$ for suitable integers $a$ and $b$.  
