Roots of $x^3+5x-18=0$ using Cardano's method Given that $x^3+5x-18=0$. We have to solve it using Cardano's method. 
Using trial $x=2$ will be a root. Dividing the equation by $x-2$ we shall get the other quadratic equation and solving that one, we shall obtain all the roots. 
But when I am trying to solve the equation using Cardan\rq s method, the calculation is becoming very difficult. I don\rq t know why. Please help. 
Here is how did I proceed. 
Let $x=u+v$. Then $x^3=u^3+v^3+3uvx$ i.e.  $x^3-3uvx-(u^3+v^3)=0$. So $-3uv=5$ and $u^3+v^3=18$. Clearly $u^3, v^3$ are the roots of 
\begin{align}
&t^2-(u^3+v^3)t+(uv)^3=0\\
\Rightarrow &t^2-18t-\frac{125}{27}=0\\
\Rightarrow &27t^2-(27\times 18)t-125=0
\end{align}
and from here when we are getting the roots of $t$, they are very complicated. Hence I do not know how to simplify them so that $x=2$ finally be achieved along with the other two roots. 
Please help me

By your work we obtain:
$$x=\sqrt[3]{9+\sqrt{81+\frac{125}{27}}}+\sqrt[3]{9-\sqrt{81+\frac{125}{27}}}=$$
$$=
\sqrt[3]{9+\frac{34}{3}\sqrt{\frac{2}{3}}}+\sqrt[3]{9+\frac{34}{3}\sqrt{\frac{2}{3}}}=\frac{1}{3}\left(\sqrt[3]{243+102\sqrt6}+\sqrt[3]{243-102\sqrt6}\right)=$$
$$=\frac{1}{3}\left(\sqrt[3]{(3+2\sqrt6)^3}+\sqrt[3]{(3-2\sqrt6)^3}\right)=\frac{1}{3}(3+2\sqrt6+3-2\sqrt6)=2.$$
