Showing $\sqrt[5]{\frac{5\sqrt5+11}{2}}+\sqrt[5]{\frac{5\sqrt5-11}{2}}=\sqrt{5} $ How do I prove the following equality?
$$\sqrt[5]{\frac{5\sqrt5+11}{2}}+\sqrt[5]{\frac{5\sqrt5-11}{2}}=\sqrt{5} $$
My approach was to notice that the first term equals the golden ratio and the second term equals the reciprocal of the golden ratio, and adding them up would give $2$ times golden ratio $- 1$, which is $\sqrt5$, but how do I show that?
 A: You can raise both sides to the $5^{th}$ power and cancel out like terms of both sides until you see that both sides look the same. That's called brute force. We can save some times off by noting some relations between the terms of the sum on the left. Specifically, put $x = \sqrt[5]{\dfrac{5\sqrt{5}+11}{2}}\implies \dfrac{1}{x} = \sqrt[5]{\dfrac{5\sqrt{5}-11}{2}}$. So we show: $x+\dfrac{1}{x} = \sqrt{5}$. Let $a = x+\dfrac{1}{x}$, then we have: $a^5 = \left(x+\dfrac{1}{x}\right)^5= x^5+\dfrac{1}{x^5}+5\left(x^3+\dfrac{1}{x^3}\right)+10\left(x+\dfrac{1}{x}\right)$. But $x^5+\dfrac{1}{x^5} = 5\sqrt{5}$, and $x^3+\dfrac{1}{x^3} = \left(x+\dfrac{1}{x}\right)^3 - 3\left(x+\dfrac{1}{x}\right)$. Substituting these identities into the equation above yields: $a^5 = 5\sqrt{5}+5(a^3-3a)+10a\implies a^5-5a^3+a-\sqrt{5} = 0$. Observe that the polynomial on the left hand side of this equation in $a$ has a factor $a - \sqrt{5}$ and it can be written as: $(a-\sqrt{5})\left(a^4 + \sqrt{5}a^3+1\right) = 0$. But $a > 0 \implies a^4+\sqrt{5}a^3 + 1  > 0$. Thus this equation implies: $a - \sqrt{5} = 0$ or $a = \sqrt{5}$. So $x+\dfrac{1}{x} = \sqrt{5}$. We're done !
A: Hint : Use Binet's formula for $n=5$ $ \displaystyle F_5 = \frac{1}{\sqrt5}((\frac{1+\sqrt5}{2})^5 - (\frac{1-\sqrt5}{2})^5) $.
