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.
 A: This screams Cardan's formula
$$
-\frac{q}{2}=2,\qquad
\frac{q^2}{4}+\frac{p^3}{27}=5
$$
Thus $q=-4$ and $p=3$. What's the real root of the following equation?
$$
x^3+3x-4=0
$$
A: I think you have meant $$\sqrt[3]{2+\sqrt5}+\sqrt[3]{2-\sqrt5}$$
Let $y=\sqrt[3]{2+\sqrt5}+\sqrt[3]{2-\sqrt5}$
$y^3=2+\sqrt5+2-\sqrt5+3(\sqrt[3]{2+\sqrt5})(\sqrt[3]{2-\sqrt5})(\sqrt[3]{2+\sqrt5}+\sqrt[3]{2-\sqrt5})=4+3\sqrt[3]{-1}y$
$$\iff y^3+3y-4=0$$ whose only real root is $1$
A: 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 $c=1$.
A: Proof: Let $$\sqrt{a+\sqrt b}=\sqrt x+\sqrt y\tag1$$
Squaring both sides, we get $$a+\sqrt b=x+y+2\sqrt{xy}\tag2$$
And from $(2)$, we see that $x+y$ must equal $a$, and $2\sqrt{xy}$ must equal $\sqrt{b}$. Thus, we have $$\begin{cases}x+y=a\\xy=\frac b4\end{cases}$$
And using substitution $x=a-y$, we get $$y(a-y)-\frac b4=0\implies y^2-ay+\frac b4=0$$
Solving should give you $y$, and since $x+y=a$, $x$ is the conjugate of $y$.

To prove $\sqrt[3]{2+\sqrt5}+\sqrt[3]{2-\sqrt5}$, another way is to set each nested radical equal to $a+b\sqrt5$ and solve. Or more specifically, we have $$\sqrt[m]{A+B\sqrt[n]{C}}=a+b\sqrt[n]{C}\tag3$$and $$\sqrt[m]{A-B\sqrt[n]C}=a-b\sqrt[n]{C}\tag4$$
A: 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
$$\sqrt[3]{2+\sqrt 5}+\sqrt[3]{2-\sqrt 5}=\sqrt[3]{\left(\frac{1+\sqrt 5}{2}\right)^3}+\sqrt[3]{\left(\frac{1-\sqrt 5}{2}\right)^3} = \frac{1+\sqrt 5}{2}+\frac{1-\sqrt 5}{2}=1$$
