How do you factor $8 x^{3} - {(2 x - y)}^{3}$ $8x^3 - (2x - y)^3$ ?

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How do you factor $8 x^{3} - {(2 x - y)}^{3}$ $8x^3 - (2x - y)^3$ ?

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Answer 1 · 2018-06-28T10:56:40

$y (12 x^{2} - 6 x y + y^{2})$ $y(12x^2-6xy+y^2)$

Explanation:

$this is a difference of cubes$ $"this is a "color(blue)"difference of cubes"$

$which factors in general as$ $"which factors in general as"$

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

$8 x^{3} = {(2 x)}^{3} \Rightarrow a = 2 x and b = 2 x - y$ $8x^3=(2x)^3rArra=2x" and "b=2x-y$

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

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

$= y (12 x^{2} - 6 x y + y^{2})$ $=y(12x^2-6xy+y^2)$

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How do you factor $8 x^{3} - {(2 x - y)}^{3}$ $8x^3 - (2x - y)^3$ ?

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How do you factor $8 x^{3} - {(2 x - y)}^{3}$ $8x^3 - (2x - y)^3$ ?