Prove that $(2+ \sqrt5)^{\frac13} + (2- \sqrt5)^{\frac13}$ is an integer When checked in calculator it is 1. But how to prove it?
Also it is not a normal addition like $x+ \frac1x$ which needs direct rationalization. So I just need something to proceed.
Since $$(1 - \sqrt{5})^3 = 16 - 8 \sqrt{5}$$ and similarly $$(1 + \sqrt{5})^3 = 16 + 8 \sqrt{5}$$ it follows that $$(2 + \sqrt{5})^{1/3} + (2 - \sqrt{5})^{1/3} = \left(\frac {16+8\sqrt{5}} 8 \right)^{1/3} +\left(\frac {16-8\sqrt{5}} 8 \right)^{1/3} = \frac{1 + \sqrt{5}}{2} + \frac{1 - \sqrt{5}}{2} = 1.$$
Not a full solution but if you use the equality: $(x+y)^3 = x^3 + y^3 + 3x^2y + 3y^2x$, you can reach the equation $z^3 + z = 4$ where $z$ is the value of the above expression.  This leads to $z = 1$ as a solution.  I cubed your formula and simplified.  I do not, however, know how to solve $z^3 + z = 4$.  Cubing both sides is safe only if $z$ is positive.