# How do you factor completely x^6 - y^6?

Using the identity ${a}^{2} - {b}^{2} = \left(a + b\right) \cdot \left(a - b\right)$ we have that

${x}^{6} - {y}^{6} = {\left({x}^{3}\right)}^{2} - {\left({y}^{3}\right)}^{2} = \left({x}^{3} - {y}^{3}\right) \cdot \left({x}^{3} + {y}^{3}\right)$

Now we know that

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

and

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

So finally is

${x}^{6} - {y}^{6} = \left(x - y\right) \cdot \left(x + y\right) \cdot \left({x}^{2} - x y + {y}^{2}\right) \cdot \left({x}^{2} + x y + {y}^{2}\right)$

Nov 21, 2015

${x}^{6} - {y}^{6} = \left(x - y\right) \left({x}^{2} + x y + {y}^{2}\right) \left(x + y\right) \left({x}^{2} - x y + {y}^{2}\right)$

#### Explanation:

The difference of squares identity can be written:

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

The difference of cubes identity can be written:

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

The sum of cubes identity can be written:

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

So:

${x}^{6} - {y}^{6} = {\left({x}^{3}\right)}^{2} - {\left({y}^{3}\right)}^{2} = \left({x}^{3} - {y}^{3}\right) \left({x}^{3} + {y}^{3}\right)$

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

If we allow Complex coefficients, then this reduces into linear factors:

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

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i = \cos \left(\frac{2 \pi}{3}\right) + \sin \left(\frac{2 \pi}{3}\right) i$ is the primitive Complex cube root of $1$.