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

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

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Answer 1 · 2015-09-23T21:30:27

Use the sum of cubes identity $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ $a^3+b^3 = (a+b)(a^2-ab+b^2)$ to find:

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

Explanation:

Use the sum of cubes identity with $a = x$ $a = x$ and $b = 2 y$ $b = 2y$ :

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

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

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

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

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