Factor each of the algebraic expressions completely. 27x^{3} - 8y^{3}

Factor each of the algebraic expressions completely. $27{x}^{3}-8{y}^{3}$
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Samantha Brown
$27{x}^{3}-8{y}^{3}$
The formula for this expression is:
${a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)$
where
${\left(3x\right)}^{3}-{\left(2y\right)}^{3}=\left(3x-2y\right)\left({\left(3x\right)}^{2}+\left(3x\right)×\left(2y\right)+{\left(2y\right)}^{2}\right)$
$=\left(3x-2y\right)\left(9{x}^{2}+6xy+4{y}^{2}\right)$
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Joseph Lewis
$27{x}^{3}-8{y}^{3}$
$={\left(3x\right)}^{3}-{\left(2y\right)}^{3}$
$=\left(3x-2y\right)\left({\left(3x\right)}^{2}+\left(3x\right)\left(2y\right)+{\left(2y\right)}^{2}\right)$
$=\left(3x-2y\right)\left({3}^{2}{x}^{2}+6xy+{2}^{2}{y}^{2}\right)$
$=\left(3x-2y\right)\left(9{x}^{2}+6xy+4{y}^{2}\right)$
$=\left(3x-2y\right)\left(9{x}^{2}+6xy+4{y}^{2}\right)$
This is factored all the way.
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RizerMix

The formula for factoring the difference of cubes is: ${u}^{3}-{v}^{3}=\left(u-v\right)\cdot \left({u}^{2}+uv+{v}^{2}\right)$
$27{x}^{3}-8{y}^{3}=\left(3x{\right)}^{3}-\left(2y{\right)}^{3}$

$\left(3x-2y\right)\cdot \left(\left(3x{\right)}^{2}+\left(3x\right)\cdot \left(2y\right)+\left(2y{\right)}^{2}\right)$
$\left(3x-2y\right)\cdot \left(9{x}^{2}+6xy+4{y}^{2}\right)$