# How do you factor 24x^6 – 1029y^3?

Jan 2, 2016

Use the difference of cubes identity to find:

$24 {x}^{6} - 1029 {y}^{3} = 3 \left(2 {x}^{2} - 7 y\right) \left(4 {x}^{4} + 14 {x}^{2} y + 49 {y}^{2}\right)$

#### Explanation:

First separate out the common scalar factor $3$:

$24 {x}^{6} - 1029 {y}^{3} = 3 \left(8 {x}^{6} - 343 {y}^{3}\right)$

Next notice that both $8 {x}^{6} = {\left(2 {x}^{2}\right)}^{3}$ and $343 {y}^{3} = {\left(7 y\right)}^{3}$ are perfect cubes.

The difference of cubes identity can be written:

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

Use this with $a = 2 {x}^{2}$ and $b = 7 y$ to find:

$8 {x}^{6} - 343 {y}^{3}$

$= {\left(2 {x}^{2}\right)}^{3} - {\left(7 y\right)}^{3}$

$= \left(2 {x}^{2} - 7 y\right) \left({\left(2 {x}^{2}\right)}^{2} + \left(2 {x}^{2}\right) \left(7 y\right) + {\left(7 y\right)}^{2}\right)$

$= \left(2 {x}^{2} - 7 y\right) \left(4 {x}^{4} + 14 {x}^{2} y + 49 {y}^{2}\right)$

Putting it all together:

$24 {x}^{6} - 1029 {y}^{3} = 3 \left(2 {x}^{2} - 7 y\right) \left(4 {x}^{4} + 14 {x}^{2} y + 49 {y}^{2}\right)$

This is as far as we can go with Real coefficients.

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

$= 3 \left(2 {x}^{2} - 7 y\right) \left(2 {x}^{2} - 7 \omega y\right) \left(2 {x}^{2} - 7 {\omega}^{2} y\right)$

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