Roots of $x^3+5x-18=0$ using Cardano's method

Question

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

Answer 1

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

Answer 2

Solving $$27t^2-27\cdot18t-125=0$$

we get $(u^3,v^3)\approx(18.25,-0.25)$

So, one of real values of $x$ will be $$\sqrt[3]{18.25}+\sqrt[3]{-0.25}\approx2$$

Answer 3

Cardano'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 $2$.