# How do you factor completely 8x^6n - 27y^9m?

Aug 30, 2016

If $m , n$ are Real constants, then:

$8 {x}^{6} n - 27 {y}^{9} m$

$= \left(2 \sqrt[3]{n} {x}^{2} - 3 \sqrt[3]{m} {y}^{3}\right) \left(4 \sqrt[3]{{n}^{2}} {x}^{4} + 6 \sqrt[3]{m n} {x}^{2} {y}^{3} + 9 \sqrt[3]{{m}^{2}} {y}^{6}\right)$

#### Explanation:

If we assume that $n , m$ are Real constants and $x , y$ are variables, then it is possible to factor this expression as a difference of cubes.

The difference of cubes identity can be written:

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

So we find:

$8 {x}^{6} n - 27 {y}^{9} m$

$= {\left(2 \sqrt[3]{n} {x}^{2}\right)}^{3} - {\left(3 \sqrt[3]{m} {y}^{3}\right)}^{3}$

$= \left(2 \sqrt[3]{n} {x}^{2} - 3 \sqrt[3]{m} {y}^{3}\right) \left({\left(2 \sqrt[3]{n} {x}^{2}\right)}^{2} + \left(2 \sqrt[3]{n} {x}^{2}\right) \left(3 \sqrt[3]{m} {y}^{3}\right) + {\left(3 \sqrt[3]{m} {y}^{3}\right)}^{2}\right)$

$= \left(2 \sqrt[3]{n} {x}^{2} - 3 \sqrt[3]{m} {y}^{3}\right) \left(4 \sqrt[3]{{n}^{2}} {x}^{4} + 6 \sqrt[3]{m n} {x}^{2} {y}^{3} + 9 \sqrt[3]{{m}^{2}} {y}^{6}\right)$