Try solving in different ways

I still prefer the old school tricks. Let $\alpha = 2-\sqrt{5}$ and $\beta = 2+\sqrt{5}$ . We will get $\alpha+\beta = 4$ and $\alpha\beta = -1$ . From the equation: $(\sqrt[3]\alpha+\sqrt[3]\beta)^3 = \alpha+\beta+3\sqrt[3]{\alpha\beta}(\sqrt[3]\alpha+\sqrt[3]\beta)$ $(\sqrt[3]\alpha+\sqrt[3]\beta)^3 = 4-(\sqrt[3]\alpha+\sqrt[3]\beta)$ Since $(\sqrt[3]\alpha+\sqrt[3]\beta)$ is real number, we solve for $\sqrt[3]\alpha+\sqrt[3]\beta = 1$

sol 1: let t=3√(2-√5) + 3√(2+√5); on cubing t^3=4-3t; t^3+3t-4=0; which has only one real root t=1, hence, 3√(2-√5) + 3√(2+√5)=1

sol 2:

we can observe that on cubing (1-√5)/2 , (1+√5)/2,we get 2-√5 ,2+√5

hence on cube rooting (1-√5)/2 , (1+√5)/2 , we get 3√(2-√5) , 3√(2+√5)

=> 3√(2-√5) =(1-√5)/2 and 3√(2+√5) =(1+√5)/2

=>3√(2-√5) + 3√(2+√5)=(1-√5)/2 + (1+√5)/2

=>3√(2-√5) + 3√(2+√5)=1