# How do you factor y^3 - 125?

Dec 21, 2015

Use the difference of cubes identity to find:

${y}^{3} - 125 = \left(y - 5\right) \left({y}^{2} + 5 y + 25\right)$

#### Explanation:

The difference of cubes identity can be written:

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

Use this with $a = y$ and $b = 5$ ...

${y}^{3} - 125$

$= {y}^{3} - {5}^{3}$

$= \left(y - 5\right) \left({y}^{2} + \left(y\right) \left(5\right) + {5}^{2}\right)$

$= \left(y - 5\right) \left({y}^{2} + 5 y + 25\right)$

The remaining quadratic factor has a negative discriminant, showing that it has no linear factors with Real coefficients.

If we are allowed Complex coefficients then we can factor a little further:

$= \left(y - 5\right) \left(y - 5 \omega\right) \left(y - 5 {\omega}^{2}\right)$

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