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

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

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Answer 1 · 2016-08-18T19:55:57

${(x - y)}^{3} + 8$ $(x-y)^3+8$

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

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

Explanation:

The sum of cubes identity can be written:

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

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

${(x - y)}^{3} + 8$ $(x-y)^3+8$

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

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

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

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

$= (x - y + 2) (x - y + 2 ω) (x - y + 2 ω^{2})$ $=(x-y+2)(x-y+2omega)(x-y+2omega^2)$

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

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

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

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

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