How do you factor $27x^3 - 8y^3$ ?

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Answer 1 · 2016-01-15T03:45:02

$27x^3-8y^3=(3x)^3-(2y)^3=color(red)((3x-2y))color(red)((9x^2+6xy+4y^2))$

$27x^3-8y^3$ fits the form of the difference of cubes, $a^3-b^3$ , where $a=3x$ and $b=2y$ .

$a^3-b^3=(3x)^3-(2y)^3$

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

Substitute the known values for $a$ and $b$ .

$(3x)^3-(2y)^3=(3x-2y)((3x)^2+(3x*2y)+(2y)^2)$

Simplify.

$(3x)^3-(2y)^3=(3x-2y)(9x^2+6xy+4y^2)$

How do you factor 27x^3 - 8y^3 27x^3 - 8y^3 ?