How to prove the equation is correct I now the Lagrange identity: $$\sqrt{a+\sqrt b}=\sqrt\frac{{a+\sqrt{a^2-b}}}{2}+\sqrt\frac{{a-\sqrt{a^2-b}}}{2}$$
but i didnt know how to prove that the equation 
$$\sqrt[3]{2+\sqrt 5}+\sqrt[3]{2-\sqrt 5}=1$$
I do not think that should be used above identity. I tried to resolve the draw as irrational but appears more complicated, so please help me to solve.
Previously, thank you for the helping.

Let $a=\sqrt[3]{2+\sqrt 5}$, $b=\sqrt[3]{2-\sqrt 5}$. By inspection $a^3+b^3=4$ and $a \cdot b=-1$.
Let $c=a+b$. Then: $$4 = a^3+b^3=(a+b)(a^2 - a \cdot b + b^2) = (a+b)\left((a+b)^2 - 3 a \cdot b\right) = c^3 + 3 c $$
But the equation $c^3 + 3 c - 4 = 0$ has $1$ as the unique real root, so $c=1$.
