How do you factor $x^{3} y^{6} + 8$ $x^3y^6 + 8$ ?

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How do you factor $x^{3} y^{6} + 8$ $x^3y^6 + 8$ ?

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Answer 1 · 2015-12-14T07:17:12

$x^{3} y^{6} + 8 = (x y^{2} + 2) (x^{2} y^{4} - 2 x y^{2} + 4)$ $x^3y^6 + 8 = (xy^2+2)(x^2y^4-2xy^2+4)$

Explanation:

The sum of cubes identity can be written:

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

Both $x^{3} y^{6} = {(x y^{2})}^{3}$ $x^3y^6 = (xy^2)^3$ and $8 = 2^{3}$ $8 = 2^3$ are perfect cubes, so let $a = x y^{2}$ $a = xy^2$ and $b = 2$ $b=2$ to find:

$x^{3} y^{6} + 8$ $x^3y^6+8$

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

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

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

That's as far as you can get with Real coefficients, but if you allow Complex coefficients this factors a little further:

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

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

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