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
 A: Well, I think we can all agree that the easiest way to prove, or to "see" that $\sqrt[5]{80\sqrt 5+176}=1+\sqrt 5$ is by using the binomial theorem.
If you're in an exam and they expect you to simplify $\sqrt[5]{80\sqrt 5+176}$ to $1+\sqrt 5$ without using a calculator, then surely the best way is by trialling different values of $m$ an $n$ for: $(m+n\sqrt{5})^5,\ $ and seeing of any of them match.
In terms of preparing for an exam where you are expected to be able to simplify like this quickly without a calculator, the only way to improve time-wise is by memorising a big chunk of Pascal's triangle. But then this exam (question) seems to me to be a bit silly, as it is requiring you to waste a chunk of your brain's memory storage without much reward. Luckily, I don't think there are many exams that have questions that require you to do this.
A: Let's rewrite our radical as $\sqrt[5]{\frac{160\sqrt{5}+352}{2}} $ to enable us to factor out $32$ from the root.
$$\Longrightarrow\sqrt[5]{32}*\sqrt[5]{\frac{5\sqrt{5}+11}{2}}$$
By equating both sides:
$$\Longrightarrow2\sqrt[5]{\frac{5\sqrt{5}+11}{2}}=1+\sqrt{5}$$
$$\Longrightarrow\sqrt[5]{\frac{5\sqrt{5}+11}{2}}=\frac{1+\sqrt{5}}{2}=\phi$$
$$\Longrightarrow\frac{5\sqrt{5}+11}{2}=\phi^5$$
Since $\phi^2=\phi+1$ we can rewrite $\phi^5$ easily as $\phi(\phi+1)^2=\left(\frac{1+\sqrt{5}}{2}\right)\left(\frac{3+\sqrt{5}}{2}\right)^2$
$$\Longrightarrow\left(\frac{1+\sqrt{5}}{2}\right)\left(\frac{7+3\sqrt{5}}{2}\right)$$
$$\Longrightarrow\frac{11+5\sqrt{5}}{2}$$
A: $$\large\sqrt[5]{80\sqrt 5+176}=\sqrt[5]{(1+\sqrt5)^5}=1+\sqrt5$$
Remember that: $$\color{red}{176+80\sqrt{5}}=(1+\sqrt5)^5$$
using a calculator to developed $(1+\sqrt5)^5$.
A: 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
$$ \sqrt[5]{175 \pm 80 \, \sqrt{5}} = 1 \pm \sqrt{5}. $$
