Express this sum of radicals as an integer? I have read somewhere that the radical $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1$ and I don't understand it. How do you solve this(when the RHS is unknown)?
 A: Let us consider the equation
$
y^3+py+q=0
$.
The theory of its solution tells, that we should consider 
$$
\Delta=\left(\frac{p}3\right)^3+\left(\frac{q}2\right)^2.
$$
If $\Delta>0$, then 
$$
y=\sqrt[3]{-\frac{q}2-\sqrt{\Delta}}+\sqrt[3]{-\frac{q}2+\sqrt{\Delta}}
$$
is unique real solution.
If you know this, the rest is simple. We guess $q=-4$, then $p=3$ and we have equation
$$
y^3+3y-4=0
$$
with an obvious real solution $y_0=1$.
A: Let $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$. Write $a=2+\sqrt{5}$ and $b=2-\sqrt{5}$. Then
$$
x^3=
a+b + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2}
=
a+b + 3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b})
=4-3x
$$
Now, the derivative of $x^3+3x-4$ is always positive, which means there is only one real root. By inspection, $x=1$ is that root.
A: Let $~\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=a.~$ Cubing both sides, using ${\underbrace{(x+y)}_a}^3=x^3+y^3+3xy~\underbrace{(x+y)}_a$, and simplifying, we have $a^3=4-3a$, whose only real solution, $a=1$, can be found quite easily via the rational root theorem. After dividing $a^3+3a-4$ by $a-1$, we're left with the quadratic $a^2+a+4$, whose roots are both complex, but $a\in$ R.
