# How do you factor 2x^9-2?

Apr 17, 2015

$2 {x}^{9} - 2$

$= 2 \left({x}^{9} - 1\right)$

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

We know that color(blue)(a^3 - b^3 = (a - b)(a^2 + ab + b^2)

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

$= 2 \left({x}^{3} - 1\right) \left({x}^{6} + {x}^{3} + 1\right)$

$= 2 \left({x}^{3} - {1}^{3}\right) \left({x}^{6} + {x}^{3} + 1\right)$

Applying the Difference of Cubes Formula again, we get

$= 2 \left(x - 1\right) \left({x}^{2} + \left(x\right) \left(1\right) + {1}^{2}\right) \left({x}^{6} + {x}^{3} + 1\right)$

color(green)( = 2(x - 1)(x^2 + x + 1)(x^6 + x^3 + 1)

As none of the Factors can be factorised further, the above becomes the factorised form of $2 {x}^{9} - 2$