Cube root sum

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

and $\large{b=\sqrt [ 3 ]{ 2-\sqrt { 5 }} }$

$\large{x= \sqrt [ 3 ]{ 2+\sqrt { 5 } } +\sqrt [ 3 ]{ 2-\sqrt { 5 } }=a+b\\ \Rightarrow x^3=(a+b)^3\\ \Rightarrow x^3=a^3+b^3+3ab(a+b)\\ \Rightarrow x^3=\left(\sqrt [ 3 ]{ 2+\sqrt { 5 } }\right)^3+\left(\sqrt [ 3 ]{ 2-\sqrt { 5 } }\right)^3+3\left(\sqrt [ 3 ]{ 2+\sqrt { 5 } }\right)\left(\sqrt [ 3 ]{ 2-\sqrt { 5 } }\right)\left(\sqrt [ 3 ]{ 2+\sqrt { 5 } }+\sqrt [ 3 ]{ 2-\sqrt { 5 } }\right)\\ \Rightarrow x^3=2+\sqrt5+2-\sqrt5+3(2^2-5)\times{x}\\ \Rightarrow x^3=4+3\times(-1)\times{x}\\ \Rightarrow x^3=4-3x\\ \Rightarrow x^3+3x-4=0}$

Solving that $\large{x=\sqrt [ 3 ]{ 2+\sqrt { 5 } } +\sqrt [ 3 ]{ 2-\sqrt { 5 } }=1}$