How do I prove that $\sqrt[5]{80\sqrt 5+176}=1+\sqrt 5$?

Question

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

Answer 1

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 $\phi(\phi+1)^2=\left(\frac{1+\sqrt{5}}{2}\right)\left(\frac{3+\sqrt{5}}{2}\right)^2$ $$\Longrightarrow\left(\frac{1+\sqrt{5}}{2}\right)\left(\frac{7+3\sqrt{5}}{2}\right)$$ $$\Longrightarrow\frac{11+5\sqrt{5}}{2}$$

Answer 2

$$\large\sqrt[5]{80\sqrt 5+176}=\sqrt[5]{(1+\sqrt5)^5}=1+\sqrt5$$

Remember that: $$\color{red}{176+80\sqrt{5}}=(1+\sqrt5)^5$$

using a calculator to developed $(1+\sqrt5)^5$.

Answer 3

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.

Answer 4

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 $$ \sqrt[5]{175 \pm 80 \, \sqrt{5}} = 1 \pm \sqrt{5}. $$