Why is $\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}} = 3$? 
How to show that $$\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}} = 3?$$

This equality comes from solving $$t^3 - 15 t - 4 = 0$$ using Cardanos fomula and knowing the solution $t_1=4$.
I have attempted multiplying the whole thing with $(\sqrt[3]{18+5\sqrt{13}})^2 - (\sqrt[3]{18-5\sqrt{13}})^2$, but no success. Then I have solved for one cubic root and put all to the third power. Also no success.

Let $(a + b\sqrt{13})^3 = (18 + 5\sqrt{13})$ for $a, b \in \Bbb Q$
Expanding the LHS gives,
$$(a^3 + 39 ab^2 - 18 ) +\sqrt{13}(3a^2 b + 13 b^3 - 5) = 0$$,
From this we get,
$$\begin{cases}a^3 + 39 ab^2 - 18 = 0 \\ 3a^2 b + 13 b^3 - 5 = 0\end{cases}$$
Solving the system give $ a = \dfrac 32$ and $ b = \dfrac12$
Therefore 
$$\sqrt[3]{(18 + 5\sqrt{13})} = \dfrac 32 +\dfrac12\sqrt{13}$$
Similarly,
$$\sqrt[3]{(18 - 5\sqrt{13})} = \dfrac 32 -\dfrac12\sqrt{13}$$
Hence the sum is $3$.
