# How do you factor 1-124x^3?

Jun 27, 2015

Use the difference of cubes identity:

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

Let $\alpha = \sqrt[3]{124}$

Then $1 - 124 {x}^{3} = \left(1 - \alpha x\right) \left(1 + \alpha x + {\alpha}^{2} {x}^{2}\right)$

#### Explanation:

I think the $124$ in the question should have been $125$, but I will answer the question as it stands.

Let $\alpha = \sqrt[3]{124}$

Then:

$1 - 124 {x}^{3}$

$= {1}^{3} - {\left(\alpha x\right)}^{3}$

$= \left(1 - \alpha x\right) \left(1 + \alpha x + {\left(\alpha x\right)}^{2}\right)$

$= \left(1 - \alpha x\right) \left(1 + \alpha x + {\alpha}^{2} {x}^{2}\right)$

The remaining quadratic factor has no simpler linear factors with real coefficients.

If the $124$ was $125 = {5}^{3}$ then we would have

$1 - 125 {x}^{3} = \left(1 - 5 x\right) \left(1 + 5 x + 25 {x}^{2}\right)$