How do you factor $((x-y)^3) +8$ ?

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Answer 1 · 2016-08-18T19:55:57

$(x-y)^3+8$

$=(x-y+2)(x^2-2xy+y^2-2x+2y+4)$

$=(x-y+2)(x-y-1+sqrt(3)i)(x-y-1-sqrt(3)i)$

The sum of cubes identity can be written:

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

Use this with $a=(x-y)$ and $b=2$ to find:

$(x-y)^3+8$

$=(x-y)^3+2^3$

$=((x-y)+2)((x-y)^2-(x-y)(2)+2^2)$

$=(x-y+2)(x^2-2xy+y^2-2x+2y+4)$

If you allow Complex coefficients, then this factors further as:

$=(x-y+2)(x-y+2omega)(x-y+2omega^2)$

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

$=(x-y+2)(x-y-1+sqrt(3)i)(x-y-1-sqrt(3)i)$

How do you factor ((x-y)^3) +8((x-y)^3) +8?