# How do you solve x^3-x=1?

Apr 24, 2017

See below.

#### Explanation:

For my approach, I will be using a graphical interpretation.

You can rewrite the equation as ${x}^{3} - x - 1 = 0$ as the first step.

Then graph the following: $f \left(x\right) = {x}^{3} - x - 1$.

graph{x^3 - x - 1 [-10, 10, -5, 5]}

Click on where the graph intersects the x-axis. This point should be (1.325, 0). Therefore, the answer is $x = 1.325$.

Apr 24, 2017

Use Cardano's method to find real root:

${x}_{1} = \frac{1}{3} \left(\sqrt{\frac{27 + 3 \sqrt{69}}{2}} + \sqrt{\frac{27 - 3 \sqrt{69}}{2}}\right)$

$\textcolor{w h i t e}{{x}_{1}} \approx 1.324717957$

and related complex roots.

#### Explanation:

Given:

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

Subtract $1$ from both sides to get:

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

$\textcolor{w h i t e}{}$
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, $a = 1$, $b = 0$, $c = - 1$ and $d = - 1$, so we find:

$\Delta = 0 + 4 + 0 - 27 + 0 = - 23$

Since $\Delta < 0$ this cubic has $1$ Real zero and $2$ non-Real Complex zeros, which are Complex conjugates of one another.

$\textcolor{w h i t e}{}$
Cardano's method

Let $x = u + v$.

Then our equation becomes:

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

Multiplying out and rearranging a little:

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

Add the constraint $v = \frac{1}{3 u}$ to eliminate the term in $\left(u + v\right)$ and get:

${u}^{3} + \frac{1}{27 {u}^{3}} - 1 = 0$

Multiply through by $27 {u}^{3}$ to get:

$27 {\left({u}^{3}\right)}^{2} - 27 \left({u}^{3}\right) + 1 = 0$

Using the quadratic formula, we find:

${u}^{3} = \frac{27 \pm \sqrt{{\left(- 27\right)}^{2} - 4 \left(27\right) \left(1\right)}}{2 \cdot 27}$

$\textcolor{w h i t e}{{u}^{3}} = \frac{27 \pm \sqrt{729 - 108}}{54}$

$\textcolor{w h i t e}{{u}^{3}} = \frac{27 \pm \sqrt{621}}{54}$

$\textcolor{w h i t e}{{u}^{3}} = \frac{27 \pm 3 \sqrt{69}}{54}$

Note that these values are both real and the derivation was symmetrical in $u , v$. So we can use one of these values for ${u}^{3}$ and the other for ${v}^{3}$ to find the real root:

${x}_{1} = \sqrt{\frac{27 + 3 \sqrt{69}}{54}} + \sqrt{\frac{27 - 3 \sqrt{69}}{54}}$

$\textcolor{w h i t e}{{x}_{1}} = \frac{1}{3} \left(\sqrt{\frac{27 + 3 \sqrt{69}}{2}} + \sqrt{\frac{27 - 3 \sqrt{69}}{2}}\right)$

$\textcolor{w h i t e}{{x}_{1}} \approx 1.324717957$

The complex roots are given by multiplying $u$ and $v$ by suitable powers of $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$, the primitive complex cube root of $1$...

${x}_{2} = \frac{1}{3} \left(\omega \sqrt{\frac{27 + 3 \sqrt{69}}{2}} + {\omega}^{2} \sqrt{\frac{27 - 3 \sqrt{69}}{2}}\right)$

${x}_{3} = \frac{1}{3} \left({\omega}^{2} \sqrt{\frac{27 + 3 \sqrt{69}}{2}} + \omega \sqrt{\frac{27 - 3 \sqrt{69}}{2}}\right)$