# How do you factor w^3 - 2y^3?

Aug 2, 2016

${w}^{3} - 2 {y}^{3} = \left(w - \sqrt[3]{2} y\right) \left({w}^{2} + \sqrt[3]{2} w y + \sqrt[3]{4} {y}^{2}\right)$

#### Explanation:

The difference of cubes identity can be written:

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

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If the coefficient of ${y}^{3}$ was a perfect cube then the example would factor naturally as a difference of two cubes.

As it is, we need to use irrational coefficients to make it into a difference of cubes:

${w}^{3} - 2 {y}^{3}$

$= {w}^{3} - {\left(\sqrt[3]{2} y\right)}^{3}$

$= \left(w - \sqrt[3]{2} y\right) \left({w}^{2} + w \left(\sqrt[3]{2} y\right) + {\left(\sqrt[3]{2} y\right)}^{2}\right)$

$= \left(w - \sqrt[3]{2} y\right) \left({w}^{2} + \sqrt[3]{2} w y + \sqrt[3]{4} {y}^{2}\right)$

The remaining quadratic factor can only be factored further with Complex coefficients, mentioned here for completeness:

${w}^{2} + \sqrt[3]{2} w y + \sqrt[3]{4} {y}^{2} = \left(w - \omega \sqrt[3]{2} y\right) \left(w - {\omega}^{2} \sqrt[3]{2} y\right)$

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