Simplifying nested radicals? $\left(\sqrt{4+\sqrt{16-a^2}}+\sqrt{4-\sqrt{16-a^2}}\right)^2$

Question

I got the following term I'd like to simplify $$\left(\sqrt{4+\sqrt{16-a^2}}+\sqrt{4-\sqrt{16-a^2}}\right)^2.$$

My Approach was to use the binomial formula. Therefore I'm currently at (hope thats correct): $$8+2×\left(\sqrt{4+\sqrt{16-a^2}}\right)×\left(\sqrt{4-\sqrt{16-a^2}}\right).$$

But now I'm stuck. Any hints or suggestions how to proceed? I had a look at denesting radicals but that doesn't help me yet.

Thanks in advance!

Answer 1

Use the fact that $\sqrt{a+b}*\sqrt{a-b}=\sqrt{(a-b)(a+b)}$, and then use @choco_addicted's comment.

Answer 2

This can be done mentally.

Expanding the square of the sum, the individual squares indeed yield $8$, and the double product is the square root of a difference of squares $\sqrt{4^2-(16-a^2)}$.

The answer is $$8+2|a|.$$