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?
 A: Hint:
$$45\pm4\sqrt{41}=(2\pm\sqrt{41})^2.$$
A: Is this the nested radical formula?
$\sqrt{45 \pm 4\sqrt{41}} = a \pm b\sqrt{41}$
$45 \pm 4\sqrt{41} = (a^2 + 41b^2) \pm 2ab\sqrt{41}$
$a^2 + 41b^ = 45; 2ab = 4 \implies a=2;b = 1$
So $\sqrt{45 \pm 4\sqrt{41}} = |2 \pm \sqrt{41}|= \pm 2 + \sqrt{41}$
Plugging that into: $=(\sqrt{45+4\sqrt{41}}-\sqrt{45-4\sqrt{41}})(45+4\sqrt{41}+45-4\sqrt{41}+(\sqrt{45+4\sqrt{41}})(\sqrt{45-4\sqrt{41}}))$
We get $(2+\sqrt{41} - (-2+\sqrt{41}))(90+(2 + \sqrt{41})(-2+\sqrt{41}))=$
$4(90 + (-4 + 41)) = 508$
===
It would have been easier not to do all the expanding:
$(2 + \sqrt{41})^3 - (-2+\sqrt{41})^3=$
$(2^3 + 3*2^2*\sqrt{41}  + 3*2*\sqrt{41}^2+\sqrt{41}^3)- (-2^3 + 3*2^2*\sqrt{41}  - 3*2*\sqrt{41}^2+\sqrt{41}^3)=$
$2(8 + 6*41) = 508$.
A: 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 $$x^2 = 90((2\cdot 4\sqrt{41})^2+1369)-2\cdot 37^3 = 258064\implies x= 508$$
A: Considering all positive values, $45\pm4\sqrt{41}$ can be written as $45\pm4\sqrt{41}=(\sqrt{41}\pm2)^2$, thus, the given expression can be simplified as follows: $$\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}=\left(\sqrt{45+4\sqrt{41}}\right)^3-\left(\sqrt{45-4\sqrt{41}}\right)^3=\left(\sqrt{\left(\sqrt{41}+2\right)^2}\right)^3-\left(\sqrt{\left(\sqrt{41}-2\right)^2}\right)^3=\left(\sqrt{41}+2\right)^3-\left(\sqrt{41}-2\right)^3$$
Now recall, $a^3-b^3=(a-b)(a^2+ab+b^2)$. Thus, $$\left(\sqrt{41}+2\right)^3-\left(\sqrt{41}-2\right)^3=\left((\sqrt{41}+2)-(\sqrt{41}-2)\right)\left((\sqrt{41}+2)^2+(\sqrt{41}+2)(\sqrt{41}-2)+(\sqrt{41}-2)^2\right)=\left(\sqrt{41}+2-\sqrt{41}+2\right)\left(45+4\sqrt{41}+41-4+45-4\sqrt{41}\right)=4(45+41-4+45)=508$$
