square root / factor problem $(A/B)^{13} - (B/A)^{13}$

This took me some time to solve. Here you go:

First, we find this:

$$\begin{aligned} (\sqrt{13+\sqrt{a}}-\sqrt{13-\sqrt{a}})^2 &=13+\sqrt{a}+13-\sqrt{a}-2\sqrt{13+\sqrt{a}}\sqrt{13-\sqrt{a}}\\ &=2(13-\sqrt{169-a}) \end{aligned}$$

So,

$$\sqrt{13+\sqrt{a}}-\sqrt{13-\sqrt{a}}=\sqrt{2}\sqrt{13-\sqrt{169-a}}$$

By what we have, we write,

$A-B=\sqrt{2}B$

What I used here is the fact that I've summed over all $a$ from $1$ to $168$, and that summing with $\sqrt{169-a}$ is the same as summing with $\sqrt{a}$ in this question.

Now, we have $A=(1+\sqrt{2})B$

Thus, $\frac{A}{B}=1+\sqrt{2}$ and $\frac{B}{A}=\sqrt{2}-1$

We just calculate $(\frac{A}{B})^{13}$ and $(\frac{B}{A})^{13}$ which I believe is okay to be done using calculator. Else, comment so I can edit my answer.

Let $$A=\sum_{n=1}^{168}\sqrt{13+\sqrt{n}},B=\sum_{n=1}^{168}\sqrt{13-\sqrt{n}}$$ since $$\sqrt{2}A=\sum_{n=1}^{168}\sqrt{26+2\sqrt{n}}=\sum_{n=1}^{168}\left(\sqrt{13+\sqrt{169-n}}+\sqrt{13-\sqrt{169-n}}\right)=A+B$$ so we have $x=\dfrac{A}{B}=\sqrt{2}$,then we have $$x=\sqrt{2}+1,\dfrac{1}{x}=\sqrt{2}-1\Longrightarrow x+\dfrac{1}{x}=2\sqrt{2}$$ let $$a_{n}=x^n-x^{-n}$$ use this well know indentity $$a_{n+2}=(x+\dfrac{1}{x})a_{n+1}-a_{n}\Longrightarrow a_{n+2}=2\sqrt{2}a_{n+1}-a_{n}$$ $$a_{1}=2,a_{2}=4\sqrt{2}$$ so $$a_{3}=2\sqrt{2}a_{2}-a_{1}=16-2=14$$ $$a_{4}=2\sqrt{2}a_{3}-a_{2}=28\sqrt{2}-4\sqrt{2}=24\sqrt{2}$$ $$a_{5}=2\sqrt{2}a_{4}-a_{3}=96-14=82$$ $$\cdots$$