# How do you factor 8y^3 - 27?

May 17, 2015

$8 {y}^{3} - 27 = {\left(2 y\right)}^{3} - {3}^{3}$

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

$= \left(2 y - 3\right) \left(4 {y}^{2} + 6 y + 9\right)$

based on the difference of cubes identity:

${m}^{3} - {n}^{3} = \left(m - n\right) \left({m}^{2} + m n + {n}^{2}\right)$,
with $m = 2 y$ and $n = 3$.

Note that $\left(4 {y}^{2} + 6 y + 9\right)$ has no simpler factors with real coefficients. To check this, note that it is of the form $a {y}^{2} + b y + c$ with $a = 4$, $b = 6$ and $c = 9$, which has discriminant given by the formula:

$\Delta = {b}^{2} - 4 a c$
$= {6}^{2} - \left(4 \times 4 \times 9\right) = 36 - 144 = - 108 < 0$

Just as a potential point of interest, if you were allowed complex coefficients, this would factor as:

$\left(4 {y}^{2} + 6 y + 9\right) = \left(2 y - 3 \omega\right) \left(2 y - 3 {\omega}^{2}\right)$

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