# How do you find all the real and complex roots of  x^6-28x^3+27=0?

Jan 22, 2016

Notice that the sum of the coefficients is zero to factor as $\left({x}^{3} - 1\right) \left({x}^{3} - 27\right)$, then identify all of the cube roots.

#### Explanation:

The primitive Complex cube root of $1$ is:

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

The cube roots of $1$ are $1$, $\omega$ and ${\omega}^{2} = \overline{\omega}$

To verify this you can factor ${x}^{3} - 1 = \left(x - 1\right) \left({x}^{2} + x + 1\right)$ then use the quadratic formula to find that the zeros of ${x}^{2} + x + 1$ are:

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

${x}^{6} - 28 {x}^{3} + 27$ is a quadratic in ${x}^{3}$ with coefficients whose sum is $0$. Hence $\left({x}^{3} - 1\right)$ is a factor and we find:

${x}^{6} - 28 {x}^{3} + 27$

$= \left({x}^{3} - 1\right) \left({x}^{3} - 27\right)$

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

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

So the roots of ${x}^{6} - 28 {x}^{3} + 27 = 0$ are:

$x = 1$

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

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

$x = 3$

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

$x = 3 {\omega}^{2} = - \frac{3}{2} - \frac{3 \sqrt{3}}{2} i$