# How do you factor z^6 - 1?

Jun 12, 2015

${z}^{6} - 1 = \left({z}^{3} - 1\right) \left({z}^{3} + 1\right)$
$= \left(z - 1\right) \left({z}^{2} + z + 1\right) \left(z + 1\right) \left({z}^{2} - z + 1\right)$

#### Explanation:

${z}^{6} - 1 = {\left({z}^{3}\right)}^{2} - {1}^{2}$ is a difference of squares, so we can use the identity:

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

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

${z}^{3} - 1 = {z}^{3} - {1}^{3}$ is a difference of cubes, so we can use the identity:

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$ to get:

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

${z}^{3} + 1 = {z}^{3} + {1}^{3}$ is a sum of cubes, so we can use the identity:

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$ to get:

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

Putting this all together we get:

${z}^{6} - 1 = \left(z - 1\right) \left({z}^{2} + z + 1\right) \left(z + 1\right) \left({z}^{2} - z + 1\right)$

If we allow complex coefficients then we can factor:

${z}^{6} - 1 = \left(z - 1\right) \left(z - \omega\right) \left(z - {\omega}^{2}\right) \left(z + 1\right) \left(z + \omega\right) \left(z + {\omega}^{2}\right)$

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

Jun 12, 2015

${z}^{6} - 1 = \left(z - 1\right) \left({z}^{2} + z + 1\right) \left(z + 1\right) \left({z}^{2} - z + 1\right)$.

#### Explanation:

Remembering the formula ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$ we can state that ${z}^{6} - 1 = \left({z}^{3} - 1\right) \left({z}^{3} + 1\right)$.

Now we have two ugly binomial to simplify. We could remember the formula of the difference of two perfect cubes , but it's not practical to memorize all formulas.

So what I suggest is to try to devide the two binomial by (z-1) and (z+1) that are the most probable divisors.
$\left({z}^{3} - 1\right) : \left(z + 1\right) \to \text{quotient"=z^2-z+1 " remainder} = - 2$
$\left({z}^{3} - 1\right) : \left(z - 1\right) \to \text{ quotient"=z^2+z+1 " remainder} = 0$
So the second division is perfect!
$\left({z}^{3} + 1\right) : \left(z + 1\right) \to \text{ quotient"=z^2-z+1 " remainder} = 0$
$\left({z}^{3} + 1\right) : \left(z - 1\right) \to \text{quotient"=z^2+z+1 " remainder} = 2$
And the first division is perfect!

Now we can rewrite the initial binomial in this way:
${z}^{6} - 1 = \left({z}^{3} - 1\right) \left({z}^{3} + 1\right) = \left(z - 1\right) \left({z}^{2} + z + 1\right) \left({z}^{3} + 1\right) = \left(z - 1\right) \left({z}^{2} + z + 1\right) \left(z + 1\right) \left({z}^{2} - z + 1\right) .$