simplify $\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}$

Question

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?

Answer 1

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:

$(2 + \sqrt{41})^3 - (-2+\sqrt{41})^3=$

$(2^3 + 3*2^2*\sqrt{41} + 3*2*\sqrt{41}^2+\sqrt{41}^3)- (-2^3 + 3*2^2*\sqrt{41} - 3*2*\sqrt{41}^2+\sqrt{41}^3)=$

$2(8 + 6*41) = 508$.

Answer 2

Hint: $$45\pm4\sqrt{41}=(2\pm\sqrt{41})^2.$$

Answer 3

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 $$x^2 = 90((2\cdot 4\sqrt{41})^2+1369)-2\cdot 37^3 = 258064\implies x= 508$$