OliForum Matematico

$ \displaystyle2+\sqrt5=(\frac{1+\sqrt5}{2})^3 $

Poniamo $t=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$. Da \begin{equation}
(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}})^3=2+\sqrt{5}+3\sqrt[3]{(2+\sqrt{5})^2(2-\sqrt{5})}+3\sqrt[3]{(2+\sqrt{5})(2-\sqrt{5})^2}+2-\sqrt{5}=\end{equation}\begin{equation}
=4+3\sqrt[3]{(2+\sqrt{5})(2-\sqrt{5})}(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}})=4-3(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}})
\end{equation}
si ricava che $t^3=4-3t$ da cui $t^3 +3t-4=0$, ma $t^3 +3t-4=(t-1)(t^2+t+4)=(t-1)[(t+1/2)^2+4-1/4]$ e quindi $t=1$.