How do you factor 1331x^3 – 8y^3?

Jan 7, 2016

Use the difference of cubes identity to find:

$1331 {x}^{3} - 8 {y}^{3} = \left(11 x - 2 y\right) \left(121 {x}^{2} + 22 x y + 4 {y}^{2}\right)$

Explanation:

Both $1331 {x}^{3} = {\left(11 x\right)}^{3}$ and $8 {y}^{3} = {\left(2 y\right)}^{3}$ are perfect cubes.

Use the difference of cubes identity, which can be written:

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

with $a = 11 x$ and $b = 2 y$ as follows:

$1331 {x}^{3} - 8 {y}^{3}$

$= {\left(11 x\right)}^{3} - {\left(2 y\right)}^{3}$

$= \left(11 x - 2 y\right) \left({\left(11 x\right)}^{2} + \left(11 x\right) \left(2 y\right) + {\left(2 y\right)}^{2}\right)$

$= \left(11 x - 2 y\right) \left(121 {x}^{2} + 22 x y + 4 {y}^{2}\right)$

The remaining quadratic factor cannot be factored into linear factors with Real coefficients (as you can tell by checking its discriminant).

You can factor it using Complex coefficients:

$= \left(11 x - 2 y\right) \left(11 x - 2 \omega y\right) \left(11 x - 2 {\omega}^{2} y\right)$

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