How do you find all the zeros of f(x)=x^3-1?

Aug 8, 2016

$f \left(x\right)$ has zeros $1$, $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ and $\overline{\omega} = - \frac{1}{2} - \frac{\sqrt{3}}{2} i$

Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

We use this below with $a = \left(x + \frac{1}{2}\right)$ and $b = \frac{\sqrt{3}}{2} i$.

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

Note that $f \left(1\right) = 1 - 1 = 0$ so $x = 1$ is a zero and $\left(x - 1\right)$ a factor...

${x}^{3} - 1$

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

$= \left(x - 1\right) \left({\left(x + \frac{1}{2}\right)}^{2} - \frac{1}{4} + 1\right)$

$= \left(x - 1\right) \left({\left(x + \frac{1}{2}\right)}^{2} + \frac{3}{4}\right)$

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

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

Hence the other two zeros are:

$x = - \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$

One of these is often denoted by the Greek letter $\omega$

$\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$

This is called the primitive Complex cube root of $1$.

The other is $\overline{\omega} = {\omega}^{2} = - \frac{1}{2} - \frac{\sqrt{3}}{2} i$