# How do you solve x^3-3x+1=0 using Cadano's method ?

May 26, 2017

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:

$\textcolor{w h i t e}{}$
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:

${\left(u + v\right)}^{3} - 3 \left(u + v\right) + 1 = 0$

which expands to:

${u}^{3} + {v}^{3} + 3 \left(u v - 1\right) \left(u + v\right) + 1 = 0$

Add the constraint $v = \frac{1}{u}$ to eliminate the $\left(u + v\right)$ 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 $\left(t - 1\right)$ 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:

${x}_{1} = \sqrt[3]{\omega} + \frac{1}{\sqrt[3]{\omega}} = 2 \cos \left(\frac{2 \pi}{9}\right) \approx 1.5321$

${x}_{2} = \omega \sqrt[3]{\omega} + \frac{1}{\omega \sqrt[3]{\omega}} = 2 \cos \left(\frac{8 \pi}{9}\right) \approx - 1.8794$

${x}_{3} = {\omega}^{2} \sqrt[3]{\omega} + \frac{1}{{\omega}^{2} \sqrt[3]{\omega}} = 2 \cos \left(\frac{14 \pi}{9}\right) \approx 0.34730$

May 27, 2017

Here's an alternative method...

#### Explanation:

Given:

${x}^{3} - 3 x + 1 = 0$

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:

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

$\textcolor{w h i t e}{0} = k \left({k}^{2} {\cos}^{3} \theta - 3 \cos \theta\right) + 1$

Putting $k = 2$ this becomes:

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

$\textcolor{w h i t e}{0} = 2 \cos 3 \theta + 1$

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:

$\theta = \pm \frac{\left(2 + 6 k\right) \pi}{9}$

So:

$x = 2 \cos \left(\pm \frac{\left(2 + 6 k\right) \pi}{9}\right)$

which results in distinct values:

${x}_{1} = 2 \cos \left(\frac{2 \pi}{9}\right) \approx 1.53209$

${x}_{2} = 2 \cos \left(\frac{8 \pi}{9}\right) \approx - 1.87939$

${x}_{3} = 2 \cos \left(\frac{14 \pi}{9}\right) \approx 0.34730$