Finding the roots of a 3rd order polynomial

Question

I have a fucntion of $x^3 -x - r=0$ for $r>0$ and I need to find its roots.

How can I do this ..?

Answer 1

One solution is given by $$x=\frac{\sqrt[3]{\sqrt{3} \sqrt{27 r^2-4}+9 r}}{\sqrt[3]{2} 3^{2/3}}+\frac{\sqrt[3]{\frac{2}{3}}}{\sqrt[3]{\sqrt {3} \sqrt{27 r^2-4}+9 r}}$$

Answer 2

An odd degree polynomial always has at least one real root, so we can be assured that there exists some $a, b, c$ such that:

\begin{align*} x^3 - x - r &= (x - a)(x^2 + bx + c) \\ &= x^3 + (b-a)x^2 + (c - ab)x - ac \end{align*}

Equating coefficients on like powers of $x$ yields:

$b-a = 0, c - ab = -1,\text{ and } ac = r$

Solving this for $a, b,\text{ and } c$ reduces the problem to factoring the quadratic $x^2 + bx + c$ (if possible).