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 ? 

It helps a lot that we know the answer in advance! Otherwise, some lucky guesswork (or knowledge of algebraic number theory) is required.
One can follow a quite similar line of reasoning to Rafael Bombelli (c.1526-1572). See Bombelli and the invention of complex numbers; or my answer to an earlier question here; or section 14.3, "Cubic Equations", of John Stillwell's book Mathematics and Its History (second edition 2002 - there is a third edition, which I don't have).
We know that if $a$ and $b$ are rational, and $(a + b\sqrt{13})^3 = c + d\sqrt{13}$, then $(a - b\sqrt{13})^3 = c - d\sqrt{13}$. So it is a natural idea - and a much less "radical" leap than Bombelli's (if you'll pardon the pun) - to look for rational $a, b$ such that $(a + b\sqrt{13})^3 = 18 + 5\sqrt{13}$. Then we will have $\sqrt[3]{18 + 5\sqrt{13}} + \sqrt[3]{18 - 5\sqrt{13}} = 2a$.
We know right away that $a = \frac{3}{2}$. (Honesty compels me to admit that I initially overlooked this obvious fact!) Equating coefficients of the irrational number $\sqrt{13}$ in the expanded version of either of the equations defining $a, b$, we have $a^3 + 39ab^2 = 18$, which simplifies to $\frac{9}{4} + 39b^2 = 12$. Taking $b > 0$, for definiteness, we have $b = \frac{1}{2}$. We now only 
need check that $3a^2b + 13b^3 = 5$, i.e. $3a^2 + 13b^2 = 10$, to see that the 
defining equations for $a, b$ are indeed satisfied.
