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\begin{document}
\title{How to Simplify $\sqrt[3]{2+\sqrt{5}}$}
\author{Amit Yaron}
\date{Jul 25, 2021}

\maketitle
We need a number Z, satisfying:

\begin{multline*}
z^{3}=2+\sqrt{5}
\end{multline*}

I am using the letter 'z' because I am going to solve it like an equation
in complex number, but with the square root of 5 instead of ``i''.

Let us take x,y such that:

\begin{multline*}
z=x+\sqrt{5}y\Rightarrow\\
\Rightarrow x+\sqrt{5}y=\sqrt[3]{2+\sqrt{5}}
\end{multline*}

Cubing both sides:

\begin{multline*}
x^{3}+3x^{2}(\sqrt{5})y+3x(5y^{2})+5\sqrt{5}y^{3}=2+\sqrt{5}\Rightarrow\\
\Rightarrow x^{3}+3\sqrt{5}x^{2}y+15xy^{2}+5\sqrt{5}y^{3}=2+\sqrt{5}
\end{multline*}

From here we can obtain a system of equations: one for the rational
part and the other for multiplications of $\sqrt{5}$:

\begin{multline*}
\begin{cases}
x^{3}+15xy^{2} & =2\\
3x^{2}y+5y^{3} & =1
\end{cases}
\end{multline*}

There are no solutions when $x=0$, so

\begin{multline*}
\frac{y}{x}\in\mathbb{R}
\end{multline*}

Let us take:

\begin{multline*}
a=\frac{y}{x}\Rightarrow\\
\Rightarrow\begin{cases}
x^{3}+15a^{2}x^{3}=2 & (1)\\
3ax^{3}+5a^{3}x^{3}=1 & (2)
\end{cases}
\end{multline*}

From both equations:

\begin{multline*}
x^{3}+15a^{2}x^{3}=2(3ax^{3}+5a^{3}x^{3})\Rightarrow\\
\Rightarrow(1+15a^{2})x^{3}=(6a+10a^{3})x^{3}
\end{multline*}

Let us divide by $x^{3}$:

\begin{multline*}
1+15a^{2}=6a+10a^{3}
\end{multline*}

From the sum of coefficient, we can see that:

\begin{multline*}
a=1
\end{multline*}

is a solution. Let us find x by plugging into equation (1):

\begin{multline*}
8x^{3}=1\Rightarrow\\
\Rightarrow x^{3}=\frac{1}{8}\Rightarrow\\
\Rightarrow x=\frac{1}{2}\Rightarrow\\
\Rightarrow y=ax=\frac{1}{2}\Rightarrow\\
\Rightarrow\sqrt[3]{2+\sqrt{5}}=x+\sqrt{5}y=\frac{1+\sqrt{5}}{2}
\end{multline*}

\end{document}
