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

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

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Answer 1 · 2018-04-25T12:38:07

$(\sqrt{x} - y) (\sqrt{x} + y) (x^{2} + x y^{2} + y^{4})$ $(sqrtx-y)(sqrtx+y)(x^2+xy^2+y^4)$

Explanation:

$this can be expressed as a difference of cubes$ $"this can be expressed as a "color(blue)"difference of cubes"$

$∙ x a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $•color(white)(x)a^3-b^3=(a-b)(a^2+ab+b^2)$

$with a = x and b = y^{2} \to {(y^{2})}^{3} = y^{6}$ $"with "a=x" and "b=y^2to(y^2)^3=y^6$

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

$we can express x - y^{2} as a difference of squares$ $"we can express "x-y^2" as a "color(blue)"difference of squares"$

$∙ x a^{2} - b^{2} = (a - b) (a + b)$ $•color(white)(x)a^2-b^2=(a-b)(a+b)$

$with a = \sqrt{x} and b = y$ $"with "a=sqrtx" and "b=y$

$= (\sqrt{x} - y) (\sqrt{x} + y)$ $=(sqrtx-y)(sqrtx+y)$

$\Rightarrow x^{3} - y^{6} = (\sqrt{x} - y) (\sqrt{x} + y) (x^{2} + x y^{2} + y^{4})$ $rArrx^3-y^6=(sqrtx-y)(sqrtx+y)(x^2+xy^2+y^4)$

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

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