How do you factor $5 u^{3} - 40 {(x + y)}^{3}$ $5u^3-40(x+y)^3$ ?

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How do you factor $5 u^{3} - 40 {(x + y)}^{3}$ $5u^3-40(x+y)^3$ ?

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Answer 1 · 2016-05-15T11:35:37

$5 u^{3} - 40 {(x + y)}^{3} = 5 (u - 2 x - 2 y) (u^{2} + 2 u x + 2 u y + 4 x^{2} + 8 x y + 4 y^{2})$ $5u^3-40(x+y)^3=5(u-2x-2y)(u^2+2ux+2uy+4x^2+8xy+4y^2)$

Explanation:

We will use the difference of cubes identity:

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

with $a = u$ $a=u$ and $b = 2 (x + y)$ $b=2(x+y)$

First separate out the common scalar factor $5$ $5$ :

$5 u^{3} - 40 {(x + y)}^{3}$ $5u^3-40(x+y)^3$

$= 5 (u^{3} - 8 {(x + y)}^{3})$ $=5(u^3-8(x+y)^3)$

$= 5 (u^{3} - {(2 (x + y))}^{3})$ $=5(u^3-(2(x+y))^3)$

$= 5 (u - 2 (x + y)) (u^{2} + u (2 (x + y)) + {(2 (x + y))}^{2})$ $=5(u-2(x+y))(u^2+u(2(x+y))+(2(x+y))^2)$

$= 5 (u - 2 x - 2 y) (u^{2} + 2 u x + 2 u y + 4 x^{2} + 8 x y + 4 y^{2})$ $=5(u-2x-2y)(u^2+2ux+2uy+4x^2+8xy+4y^2)$

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How do you factor $5 u^{3} - 40 {(x + y)}^{3}$ $5u^3-40(x+y)^3$ ?

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