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

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Answer 1 · 2016-01-10T21:09:29

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

You will find that $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $a^3-b^3 = (a-b)(a^2+ab+b^2)$

Therefore by setting $a = x^{2}$ $a=x^2$ and $b = 2 y$ $b=2y$ :

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

Don't think it does but it possibly could be factorised more.

How do you factor x^6 - 8y^3x6−8y3x^6 - 8y^3?