# How do you factor x^3 - 3x^2 - 1 = 0?

Jun 3, 2015

$f \left(x\right) = {x}^{3} - 3 {x}^{2} - 1$

graph{x^3-3x^2-1 [-20, 20, -10, 10]}

By the rational roots theorem, any rational roots of $f \left(x\right)$ would be $\pm 1$ and neither is a root of $f \left(x\right) = 0$ :

$f \left(- 1\right) = - 1 - 3 - 1 = - 5$

$f \left(1\right) = 1 - 3 - 1 = - 3$.

I suspect there's a missing $+ 3 x$ term in the question, but I'll have a go at the question as it stands.

First, let $y = x - 1$

${y}^{3} = {\left(x - 1\right)}^{3} = {x}^{3} - 3 {x}^{2} + 3 x - 1$

$- 3 y = - 3 \left(x - 1\right) = - 3 x + 3$

$- 3 = - 3$

So: ${y}^{3} - 3 y - 3 = {x}^{3} - 3 {x}^{2} - 1 = f \left(x\right)$

Next, using Cardano's method, let $y = u + v$

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

$= {u}^{3} + 3 {u}^{2} v + 3 u {v}^{2} + {v}^{3} - 3 \left(u + v\right) - 3$

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

$= {u}^{3} - 3 + {v}^{3}$ (if $v = \frac{1}{u}$)

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

Multiply both ends by ${u}^{3}$ to get:

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

So ${u}^{3} = \frac{3 \pm \sqrt{5}}{2}$

Since this derivation is symmetric in $u$ and $v$, we can put

${u}^{3} = \frac{3 + \sqrt{5}}{2}$

${v}^{3} = \frac{3 - \sqrt{5}}{2}$

So there's a real root:

$y = u + v = \sqrt[3]{\frac{3 + \sqrt{5}}{2}} + \sqrt[3]{\frac{3 - \sqrt{5}}{2}}$

So

${x}_{1} = y + 1 = 1 + \sqrt[3]{\frac{3 + \sqrt{5}}{2}} + \sqrt[3]{\frac{3 - \sqrt{5}}{2}}$

The other two roots are complex (!) :

${x}_{2} = 1 + \omega \sqrt[3]{\frac{3 + \sqrt{5}}{2}} + {\omega}^{2} \sqrt[3]{\frac{3 - \sqrt{5}}{2}}$

${x}_{3} = 1 + {\omega}^{2} \sqrt[3]{\frac{3 + \sqrt{5}}{2}} + \omega \sqrt[3]{\frac{3 - \sqrt{5}}{2}}$

where $\omega = - \frac{1}{2} + i \frac{\sqrt{3}}{2}$ is the primitive cube root of unity.

$f \left(x\right) = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right) =$ something horribly complicated.

I'm pretty sure this is more complex than you're supposed to know about and my suspicion about the missing term in the question is justified.