Prove that $(2+ \sqrt5)^{\frac13} + (2- \sqrt5)^{\frac13}$ is an integer

Since $$(1 - \sqrt{5})^3 = 16 - 8 \sqrt{5}$$ and similarly $$(1 + \sqrt{5})^3 = 16 + 8 \sqrt{5}$$ it follows that $$(2 + \sqrt{5})^{1/3} + (2 - \sqrt{5})^{1/3} = \left(\frac {16+8\sqrt{5}} 8 \right)^{1/3} +\left(\frac {16-8\sqrt{5}} 8 \right)^{1/3} = \frac{1 + \sqrt{5}}{2} + \frac{1 - \sqrt{5}}{2} = 1.$$

Use the general formula $x^3+y^3=(x+y)\left((x+y)^2-3xy\right)$.

Here $x=\sqrt[3]{2+\sqrt{5}}$ and $y=\sqrt[3]{2-\sqrt{5}}$.

Let $a=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$ and

$$b=\sqrt[3]{2+\sqrt{5}}\sqrt[3]{2-\sqrt{5}}=\sqrt[3]{2^2-(\sqrt{5})^2}=-1$$

$$(2+\sqrt{5})+(2-\sqrt{5})=4$$

$$=a\left(a^2-3b\right)=a^3+3a$$

$$a^3+3a-4=0$$

You can apply Rational Root Theorem/Test, polynomial division, fundamental theorem of algebra, etc.

$$(a-1)\left(a^2+a+4\right)=0$$

$a^2+a+4=0$ has no real solutions, but clearly $a$ is real, so $a=1$.

We have $x- \left(2+\sqrt 5 \right)^{\frac{1}{3}} - \left(2-\sqrt 5 \right)^{\frac{1}{3}} = 0$

So $x^3 -(2-\sqrt 5)-(2+\sqrt 5) = 3x \left(2+\sqrt 5 \right)^{\frac{1}{3}}\left(2-\sqrt 5 \right)^{\frac{1}{3}}= -3x$

(invoking that $a^3+b^3+c^3 = 3abc$ when $a+b+c = 0$)

from which we see that $x$ is a root of $x^3+3x-4 =0$

Since the derivative is positive, this means it has only one real root, which by inspection is $x=1$