# How do you factor a^9 + b^12 c^15?

Dec 5, 2015

${a}^{9} + {b}^{12} {c}^{15} = \left({a}^{3} + {b}^{4} {c}^{5}\right) \left({a}^{6} - {a}^{3} {b}^{4} {c}^{5} + {b}^{8} {c}^{10}\right)$

#### Explanation:

${a}^{9} + {b}^{12} {c}^{15}$ is of the form ${A}^{3} + {B}^{3}$ where $A = {a}^{3}$ and $B = {b}^{4} {c}^{5}$.

The sum of cubes identity tells us that:

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

Hence:

${a}^{9} + {b}^{12} {c}^{15} = {\left({a}^{3}\right)}^{3} + {\left({b}^{4} {c}^{5}\right)}^{3}$

$= \left({a}^{3} + {b}^{4} {c}^{5}\right) \left({\left({a}^{3}\right)}^{2} - \left({a}^{3}\right) \left({b}^{4} {c}^{5}\right) + {\left({b}^{4} {c}^{5}\right)}^{2}\right)$

$= \left({a}^{3} + {b}^{4} {c}^{5}\right) \left({a}^{6} - {a}^{3} {b}^{4} {c}^{5} + {b}^{8} {c}^{10}\right)$

If we allow Complex coefficients then this can be factored further as:

$= \left({a}^{3} + {b}^{4} {c}^{5}\right) \left({a}^{3} + \omega {b}^{4} {c}^{5}\right) \left({a}^{3} + {\omega}^{2} {b}^{4} {c}^{5}\right)$

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