# How do you factor 81r^3 - 3t^3?

May 9, 2015

$81 {r}^{3} - 3 {t}^{3} = 0$ has solutions when ${\left(\frac{t}{r}\right)}^{3} = 27$, so $\frac{t}{r} = 3 , 3 \omega \mathmr{and} 3 {\omega}^{2}$, where $\omega$ is the complex cube root of $1$, so $\left(3 r - t\right)$, $\left(3 \omega r - t\right)$ and $\left(3 {\omega}^{2} r - t\right)$ are all factors of $81 {r}^{3} - 3 t$.

If we restrict ourselves to real coefficients, the only linear factor is $\left(3 r - t\right)$. The other (quadratic) factor will be of the form $\left(a {r}^{2} + b r t + c {t}^{2}\right)$. We could calculate this using $\omega$, but let's stick to real numbers for simplicity...

Let's try multiplying it out:
(3r - t)(ar^2 + brt + ct^2) = 3ar^3 + (3b - a)r^2t + (3c - b)rt^2 -ct^3

Comparing the coefficients with the original polynomial, we get:
$3 a = 81$
$3 b - a = 0$
$3 c - b = 0$
$c = 3$

Hence $a = 27$, $b = 9$ and $c = 3$.

So $81 {r}^{3} - 3 {t}^{3} = \left(3 r - t\right) \left(27 {r}^{2} + 9 r t + 3 {t}^{2}\right) = 3 \left(3 r - t\right) \left(9 {r}^{2} + 3 r t + {t}^{2}\right)$.