<s>[INST] 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.
 [/INST] Observe that
$$\left(\frac{1+\sqrt 5}{2}\right)^3=\frac{1+3\sqrt 5+3\cdot5+5\sqrt 5}{8} = \frac{16+8\sqrt 5}{8}=2+\sqrt 5$$
Similarly,
$$\left(\frac{1-\sqrt 5}{2}\right)^3=\frac{1-3\sqrt 5+3\cdot5-5\sqrt 5}{8} = \frac{16-8\sqrt 5}{8}=2-\sqrt 5$$
Your equation transforms into
$$\sqrt[3]{2+\sqrt 5}+\sqrt[3]{2-\sqrt 5}=\sqrt[3]{\left(\frac{1+\sqrt 5}{2}\right)^3}+\sqrt[3]{\left(\frac{1-\sqrt 5}{2}\right)^3} = \frac{1+\sqrt 5}{2}+\frac{1-\sqrt 5}{2}=1$$
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