Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$

$$x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\\x^3=2+\sqrt5+2-\sqrt5+3\sqrt[3]{2+\sqrt{5}}\cdot\sqrt[3]{2-\sqrt{5}}(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}})\\x^3=4+3\cdot(-1)\cdot(x)$$so $$x^3+3x-4=0 \\(x-1)(x^2+x+4)\to\\ x=1,x^2+x+4=0 ,\Delta <0\\x=1$$

By observing that $$\left(\frac12\pm\frac{\sqrt 5}{2}\right)^3=\frac18\pm\frac{3\sqrt 5}8+\frac{15}{8}\pm\frac{5\sqrt 5}{8}=2\pm\sqrt 5$$ It follows $$\sqrt[3]{2+\sqrt 5}+\sqrt[3]{2-\sqrt 5}=\tfrac12+\tfrac{\sqrt 5}{2}+\tfrac12-\tfrac{\sqrt 5}2=1$$

Let $a=\sqrt[3]{2+\sqrt{5}}, b=\sqrt[3]{2-\sqrt{5}}\,$, then:

$$\require{cancel} a^3+b^3 = 2+\cancel{\sqrt{5}} + 2-\cancel{\sqrt{5}} = 4 \\ ab = \sqrt[3]{(2+\sqrt{5})(2-\sqrt{5})} = \sqrt[3]{2^2 - 5} = -1 $$

From $\,(a+b)^3 = a^3+b^3+3ab(a+b)\,$ and given that $\,x=a+b\,$ it follows that $\,x^3=4-3x\,$ $\iff 0 = x^3+3x-4=(x-1)(x^2+x+4)\,$, where only the first factor has a real root.