# How do you factor 8x^6 y^9 - 64a^3 b^12?

Oct 31, 2017

$8 {x}^{6} {y}^{9} - 64 {a}^{3} {b}^{12} = 8 \left({x}^{2} {y}^{3} - 2 a {b}^{4}\right) \left({x}^{4} {y}^{6} + 2 {x}^{2} {y}^{3} a {b}^{4} + 4 {a}^{2} {b}^{8}\right)$

#### Explanation:

Note that both terms are divisible by $8$, so we can separate that out as a factor first:

$8 {x}^{6} {y}^{9} - 64 {a}^{3} {b}^{12} = 8 \left({x}^{6} {y}^{9} - 8 {a}^{3} {b}^{12}\right)$

Note that both ${x}^{6} {y}^{9} = {\left({x}^{2} {y}^{3}\right)}^{3}$ and $8 {a}^{3} {b}^{12} = {\left(2 a {b}^{4}\right)}^{3}$ are perfect cubes.

Also note that:

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

So we can use this "difference of cubes identity" with $A = {x}^{2} {y}^{3}$ and $B = 2 a {b}^{4}$ to find:

${x}^{6} {y}^{9} - 8 {a}^{3} {b}^{12}$

$= {\left({x}^{2} {y}^{3}\right)}^{3} - {\left(2 a {b}^{4}\right)}^{3}$

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

$= \left({x}^{2} {y}^{3} - 2 a {b}^{4}\right) \left({x}^{4} {y}^{6} + 2 {x}^{2} {y}^{3} a {b}^{4} + 4 {a}^{2} {b}^{8}\right)$

Neither of the remaining factors has simpler factorisations using real coefficients. It is possible to factor the second decic expression using complex coefficients, most naturally using the primitive complex cube root of $1$:

$\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$

Then:

${x}^{4} {y}^{6} + 2 {x}^{2} {y}^{3} a {b}^{4} + 4 {a}^{2} {b}^{8} = \left({x}^{2} {y}^{3} - 2 \omega a {b}^{4}\right) \left({x}^{2} {y}^{3} - 2 {\omega}^{2} a {b}^{4}\right)$