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

Cardono's formula says that one root is$$\sqrt[3]{9+\sqrt{81+\frac{125}{27}}}+\sqrt[3]{9-\sqrt{81+\frac{125}{27}}}=\sqrt[3]{9+\frac{34}3\sqrt{\frac23}}+\sqrt[3]{9-\frac{34}3\sqrt{\frac23}}.$$But$$\sqrt[3]{9+\frac{34}3\sqrt{\frac23}}=1+2\sqrt{\frac23}\text{ and }\sqrt[3]{9-\frac{34}3\sqrt{\frac23}}=1-2\sqrt{\frac23}.$$Adding these two numbers, you'll get $2$.
