How do you factor 1/8 + 8x^3?

Dec 19, 2015

Use the sum of cubes identity to find:

$\frac{1}{8} + 8 {x}^{3} = \left(\frac{1}{2} + 2 x\right) \left(\frac{1}{4} - x + 4 {x}^{2}\right)$

Explanation:

The sum of cubes identity can be written:

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

Both $\frac{1}{8} = {\left(\frac{1}{2}\right)}^{3}$ and $8 {x}^{3} = {\left(2 x\right)}^{3}$ are perfect cubes, so it is natural to use the sum of cubes identity to help factor this, putting $a = \frac{1}{2}$ and $b = 2 x$ ...

$\frac{1}{8} + 8 {x}^{3}$

$= {\left(\frac{1}{2}\right)}^{3} + {\left(2 x\right)}^{3}$

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

$= \left(\frac{1}{2} + 2 x\right) \left(\frac{1}{4} - x + 4 {x}^{2}\right)$