# How do you factor 64y^3 + 27?

May 22, 2015

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

$64 {y}^{3} + 27 = \left({\left(4 y\right)}^{3} + {3}^{3}\right)$

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

$= \left(4 y + 3\right) \left(16 {y}^{2} - 12 y + 9\right)$

The factor $\left(16 {y}^{2} - 12 y + 9\right)$ has no linear factors with real coefficients:

The discriminant

$\Delta \left(16 {y}^{2} - 12 y + 9\right) = {\left(- 12\right)}^{2} - \left(4 \times 16 \times 9\right)$
$= 144 - 576 = - 432 < 0$

...so $16 {y}^{2} - 12 y + 9 = 0$ has no real roots.

If you are curious, the other (complex) linear factors are actually

$\left(4 y + 3 \omega\right) \left(4 y + 3 {\omega}^{2}\right)$, where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$