How do I prove that $\sqrt[5]{80\sqrt 5+176}=1+\sqrt 5$? How do I prove that $\sqrt[5]{80\sqrt 5+176}=1+\sqrt{5}$?
I have no idea how to proceed except just raising both sides to the power of $5$ and expanding $ \left(1+\sqrt{5}\right)^{5}$ using the binomial theorem

Consider: $\sqrt[5]{176 + 80 \, \sqrt{5}}$. Both integer values are multiples of $16$, or $2^4$, and leads to $\sqrt[5]{2^4 \, (11 + 5 \sqrt{5})}$. Now,
\begin{align}
11 + 5 \, \sqrt{5} &= 6 + 5 \, \sqrt{5} + 5 = (1 + 2 \, \sqrt{5} + 5) + 5 + 3 \, \sqrt{5} \\
&= (1 + \sqrt{5})^2 + \sqrt{5} \, (3 + \sqrt{5}) \\
&= (1 + \sqrt{5})^2 \, \left(1 + \frac{\sqrt{5}}{2}\right) \\
&= \frac{(1+\sqrt{5})^2 \, (2 + \sqrt{5})}{2}.  
\end{align}
Using $$2 + \sqrt{5} = \frac{16 + 8 \, \sqrt{5}}{8} = \frac{1 + 3 \, \sqrt{5} + 3 \,(\sqrt{5})^2 + 5 \, \sqrt{5}}{8} = \frac{(1+\sqrt{5})^3}{2^3} $$
then
$$ 11 + 5 \, \sqrt{5} = \frac{(1+\sqrt{5})^5}{2^4} $$
and
$$ \sqrt[5]{176 + 80 \, \sqrt{5}} = \sqrt[5]{2^4 \, \frac{(1+\sqrt{5})^5}{2^4}} = \sqrt[5]{(1 + \sqrt{5})^5} = 1 + \sqrt{5}.$$
Following the same pattern one can determine that $\sqrt[5]{176 - 80 \, \sqrt{5}} = 1 - \sqrt{5}$ which leads to the statement
$$ \sqrt[5]{175 \pm 80 \, \sqrt{5}} = 1 \pm \sqrt{5}. $$
