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

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Answer 1 · 2016-03-03T21:12:25

$x^3+1/8=(x+1/2)(x^2-1/2x+1/4)$

Both $x^3$ and $1/8 = (1/2)^3$ are perfect cubes. So we can use the "sum of cubes" identity:

$a^3+b^3 = (a+b)(a^2-ab+b^2)$

with $a=x$ and $b=1/2$ as follows...

$x^3+1/8$

$=x^3+(1/2)^3$

$=(x+1/2)(x^2-x(1/2)+(1/2)^2)$

$=(x+1/2)(x^2-1/2x+1/4)$

The remaining quadratic factor has no simpler factors with Real coefficients, but we can factor it if we allow Complex coefficients:

$=(x+1/2)(x+1/2omega)(x+1/2omega^2)$

where $omega = -1/2+sqrt(3)/2i$ is the primitive Complex cube root of $1$ .

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