How to prove that $(-18+\sqrt{325})^{\frac{1}{3}}+(-18-\sqrt{325})^{\frac{1}{3}} = 3$ How to prove that $\left(-18+\sqrt{325}\right)^{\frac{1}{3}}+\left(-18-\sqrt{325}\right)^{\frac{1}{3}} = 3$ in a direct way ?
I have found one indirect way to do so: Define $t=\left(-18+\sqrt{325}\right)^{\frac{1}{3}}+\left(-18-\sqrt{325}\right)^{\frac{1}{3}}$, and then observe $t$ is a root of $f(x) = x^3 + 3x - 36 = 0$. Also observe that $f(3) = 0$. Now $f(x)$ has only one real root, since $f'(x) = x^2+3$ has no real solution. So $t =3$, as desired. 
But I couldn't find any direct/algebraic way of fiddling it for a while. I guess morever this is true:  $\left(-\frac{t(t^2+3)}{2}+\sqrt{\left[\frac{t(t^2+3)}{2}\right]^2 + 1}\right)^{\frac{1}{3}}+\left(-\frac{t(t^2+3)}{2}-\sqrt{\left[\frac{t(t^2+3)}{2}\right]^2 + 1}\right)^{\frac{1}{3}} = t$, which can possibly be proved in the same indirect way but I don't know how to prove this by direct manipulation. 
One way maybe is to write $z_+ = \left(-\frac{t(t^2+3)}{2}+\sqrt{\left[\frac{t(t^2+3)}{2}\right]^2 + 1}\right)^{\frac{1}{3}}$, $z_- = \left(-\frac{t(t^2+3)}{2}-\sqrt{\left[\frac{t(t^2+3)}{2}\right]^2 + 1}\right)^{\frac{1}{3}}$, and writing $t = z_+ + z_-$, observe $z_+z_- = -1$, so $(x-z_+)(x-z_-) = x^2 - tx -1$, and then with Fundamental Theorem of Symmetric Polynomials, maybe calculating $(z_+^3 - z_-^3)^2$ and $(z_+^3 + z_-^3)$ in terms of $t$ would help ? 

Compute
$$\left(\frac{3+\sqrt{13}}2\right)^3=18+5\sqrt{13}.$$
Therefore
$$\sqrt[3]{18+5\sqrt{13}}=\frac{3+\sqrt{13}}2.$$
Similarly
$$\sqrt[3]{\pm 18\pm 5\sqrt{13}}=\frac{\pm 3\pm\sqrt{13}}2$$
where the signs on both sides correspond. Then
$$\sqrt[3]{-18+ 5\sqrt{13}}-\sqrt[3]{-18-5\sqrt{13}}
=\frac{-3+\sqrt{13}}2+\frac{-3-\sqrt{13}}2=-3.$$
