# How do you factor 1 - 256y^8?

Jul 15, 2015

Use the difference of squares identity three times to find:

$1 - 256 {y}^{8} = \left(1 - 2 y\right) \left(1 + 2 y\right) \left(1 + 4 {y}^{2}\right) \left(1 + 16 {y}^{4}\right)$

#### Explanation:

$1 - 256 {y}^{8}$

$= {1}^{2} - {\left(16 {y}^{4}\right)}^{2}$

$= \left(1 - 16 {y}^{4}\right) \left(1 + 16 {y}^{4}\right)$

$= \left({1}^{2} - {\left(4 {y}^{2}\right)}^{2}\right) \left(1 + 16 {y}^{4}\right)$

$= \left(1 - 4 {y}^{2}\right) \left(1 + 4 {y}^{2}\right) \left(1 + 16 {y}^{4}\right)$

$= \left({1}^{2} - {\left(2 y\right)}^{2}\right) \left(1 + 4 {y}^{2}\right) \left(1 + 16 {y}^{4}\right)$

$= \left(1 - 2 y\right) \left(1 + 2 y\right) \left(1 + 4 {y}^{2}\right) \left(1 + 16 {y}^{4}\right)$

...using the difference of squares identity:

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