# How do you factor x^6+2x^3+1 ?

Jul 1, 2015

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

#### Explanation:

${x}^{6} + 2 {x}^{3} + 1$

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

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

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

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

Using the sum of cubes identity:

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

with $a = x$ and $b = 1$

$\left({x}^{2} - x + 1\right)$ has no simpler factors with real coefficients.