<s>[INST] simplify $\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}$ simplify $\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}$.

1.$90^{\frac{3}{2}}$
2.$106\sqrt{41}$
3.$4\sqrt{41}$
4.$504$
5.$508$

My attempt:I do like this but I didn't get any of those five.
$\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}={\sqrt{45+4\sqrt{41}}}^3-\sqrt{45-4\sqrt{41}}^3$
$=(\sqrt{45+4\sqrt{41}}-\sqrt{45-4\sqrt{41}})(45+4\sqrt{41}+45-4\sqrt{41}+(\sqrt{45+4\sqrt{41}})(\sqrt{45-4\sqrt{41}})$
Now I do the nested radicals formula and I get $254\sqrt{41}$ which is none of those where did I mistaked?
 [/INST] When I see expression where both $\alpha = a+b\sqrt{n}$ and $\beta =a-b\sqrt n$ occur, I immediately calculate $\alpha + \beta = 2a$ and $\alpha\beta = a^2-nb^2$ since they are guaranteed to be integers (more precisely, the minimal polynomial of both of them is $x^2 - (\alpha+\beta)x+\alpha\beta$ which might be helpful in some cases; if you are not familiar with the term, just ignore this remark). So, we have $$x = \sqrt{\alpha^3}-\sqrt{\beta^3}\implies x^2 = \alpha^3 -2\sqrt{(\alpha\beta)^3}+\beta^3\\
\implies x^2 = (\alpha + \beta)(\alpha^2-\alpha\beta+\beta^2)-2\sqrt{(\alpha\beta)^3}\\
\implies x^2 = (\alpha + \beta)((\alpha-\beta)^2+\alpha\beta)-2\sqrt{(\alpha\beta)^3}$$
Now, in your case $\alpha\beta = 1369 = 37^2$, so we have $$x^2 = 90((2\cdot 4\sqrt{41})^2+1369)-2\cdot 37^3 = 258064\implies x= 508$$
</s>