How to simplify $\sqrt{\sqrt{5}+1} \cdot \sqrt{\sqrt{5}-1}$?

Question

This is the original problem: $\sqrt{\sqrt{5}+1} \cdot \sqrt{\sqrt{5}-1} = x$.

I'm really confused about how to solve this problem, I come as far as saying this: $\sqrt[4]{5} + \sqrt{1}\cdot \sqrt[4]{5}-\sqrt{1}$.

Answer 1

Hint: Start with $\sqrt{\sqrt{5}+1} \cdot \sqrt{\sqrt{5}-1} = \sqrt{(\sqrt{5}+1) \cdot (\sqrt{5}-1)}$, and then use the difference of squares identity, i.e. $(a+b)(a-b) = a^2-b^2$.

Answer 2

$$\sqrt{\sqrt{5}+1}\sqrt{\sqrt{5}-1}=\sqrt{(\sqrt{5}+1)(\sqrt{5}-1)}$$ $$=\sqrt{5-\sqrt{5}+\sqrt{5}-1}$$ $$=\sqrt{5-1}$$ $$=\sqrt{4}$$ $$=2$$