## How do you solve \(x^{3}-3 x+1=0\) using Cadano’s method?

Answer 1:

See explanation…

Explanation:

Assuming you mean Cardano’s method, I would remark that it is not a good choice for this particular cubic. This cubic has 3 real zeros, as we can find by examining its discriminant:

Discriminant

The discriminant \(\Delta\) of a cubic polynomial in the form \(a x^{3}+b x^{2}+c x+d\) is given by the formula:

\(\Delta=b^{2} c^{2}-4 a c^{3}-4 b^{3} d-27 a^{2} d^{2}+18 a b c d\)In our example, \(a=1, b=0, c=-3\) and \(d=1\), so we find:

\(\Delta=0+108+0-27+0=81\)Since \(\Delta>0\) this cubic has \(3\) Real zeros.

As a result, if we attempt to solve using Cardano’s method then the solution will be expressed in terms of irreducible cube roots of complex numbers – Cardano’s “casus irreducibilis”.

Let’s go ahead anyway to see it happen…

Given:

\(x^{3}-3 x+1=0\)Let \(x=u+v\)

Then:

\((u+v)^{3}-3(u+v)+1=0\)Add the constraint \(v=\frac{1}{u}\) to eliminate the (u+v) term and get:

\(u^{3}+\frac{1}{u^{3}}+1=0\)Multiply through by \(u^{3}\) and rearrange slightly to get:

\(\left(u^{3}\right)^{2}+\left(u^{3}\right)+1=0\)Note that this is of the form \(t^{2}+t+1=0\) which if multiplied by \((t-1)\) gives \(t^{3}-1=0\). In other words, its solutions are the non-real complex cube roots of 1:

\(\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} i=\cos \left(\frac{2 \pi}{3}\right)+i \sin \left(\frac{2 \pi}{3}\right)\)

\(\omega^{2}=-\frac{1}{2}-\frac{\sqrt{3}}{2} i=\cos \left(-\frac{2 \pi}{3}\right)+i \sin \left(-\frac{2 \pi}{3}\right)\)

Then using \(x=u+v\) and \(v=\frac{1}{u}\), we find solutions of our cubic:

Answer 2:

Here’s an alternative method…

Explanation:

Given:

Here’s an alternative trigonometric sustitution method of solution, suitable for cubics such as this one, with three real zeros (Cardano’s “casus irreducibilis”):

Consider the substitution:

\(x=k \cos \theta\)Then our cubic equation becomes:

\(0=(k \cos \theta)^{3}-3(k \cos \theta)+1\)

\(=k\left(k^{2} \cos ^{3} \theta-3 \cos \theta\right)+1\)

Putting \(k=2\) this becomes:

\(\begin{aligned}0 &=2\left(4 \cos ^{3} \theta-3 \cos \theta\right)+1 \\

&=2 \cos 3 \theta+1

\end{aligned}\)

Hence:

\(\cos 3 \theta=-\frac{1}{2}\)Hence:

\(3 \theta=\pm \cos ^{-1}\left(-\frac{1}{2}\right)+2 k \pi=\pm \frac{2 \pi}{3}+2 k \pi\)for any integer \(k\).

So:

\(x=2 \cos \left(\pm \frac{(2+6 k) \pi}{9}\right)\)which results in distinct values: