# How do you factor 8x^6 - 27y^6?

$8 {x}^{6} - 27 {y}^{6} = {2}^{3} {\left({x}^{2}\right)}^{3} - {3}^{3} {\left({y}^{2}\right)}^{3} = {\left(2 {x}^{2}\right)}^{3} - {\left(3 {y}^{2}\right)}^{3} = \left(2 {x}^{2} - 3 {y}^{2}\right) \left(4 {x}^{4} + 6 {x}^{2} {y}^{2} + 9 {y}^{4}\right)$

#### Explanation:

What jumps out at me is that everything in this expression can be expressed in terms of cubes:

$8 {x}^{6} - 27 {y}^{6} = {2}^{3} {\left({x}^{2}\right)}^{3} - {3}^{3} {\left({y}^{2}\right)}^{3} = {\left(2 {x}^{2}\right)}^{3} - {\left(3 {y}^{2}\right)}^{3}$

and, in fact, we can factor this using the general formula:

${A}^{3} - {B}^{3} = \left(A - B\right) \left({A}^{2} + A B + {B}^{2}\right)$

so let's do that.

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

May 4, 2017

$8 {x}^{6} - 27 {y}^{6}$

$= \left(\sqrt{2} x - \sqrt{3} y\right) \left(2 {x}^{2} + \sqrt{6} x y + 3 {y}^{2}\right) \left(\sqrt{2} x + \sqrt{3} y\right) \left(2 {x}^{2} - \sqrt{6} x y + 3 {y}^{2}\right)$

#### Explanation:

If we allow irrational coefficients, then this sextic expression will factor as far as a mixture of linear and quadratic factors.

Use the following identities:

Difference of squares:

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

Difference of cubes:

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

Sum of cubes:

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

Using the difference of squares:

$8 {x}^{6} - 27 {y}^{6} = {\left(2 \sqrt{2} {x}^{3}\right)}^{2} - {\left(3 \sqrt{3} {y}^{3}\right)}^{2}$

$\textcolor{w h i t e}{8 {x}^{6} - 27 {y}^{6}} = \left(2 \sqrt{2} {x}^{3} - 3 \sqrt{3} {y}^{3}\right) \left(2 \sqrt{2} {x}^{3} + 3 \sqrt{3} {y}^{3}\right)$

$\textcolor{w h i t e}{8 {x}^{6} - 27 {y}^{6}} = \left({\left(\sqrt{2} x\right)}^{3} - {\left(\sqrt{3} y\right)}^{3}\right) \left({\left(\sqrt{2} x\right)}^{3} + {\left(\sqrt{3} y\right)}^{3}\right)$

Using the difference of cubes:

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

$\textcolor{w h i t e}{{\left(\sqrt{2} x\right)}^{3} - {\left(\sqrt{3} y\right)}^{3}} = \left(\sqrt{2} x - \sqrt{3} y\right) \left(2 {x}^{2} + \sqrt{6} x y + 3 {y}^{2}\right)$

Using the sum of cubes:

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

$\textcolor{w h i t e}{{\left(\sqrt{2} x\right)}^{3} + {\left(\sqrt{3} y\right)}^{3}} = \left(\sqrt{2} x + \sqrt{3} y\right) \left(2 {x}^{2} - \sqrt{6} x y + 3 {y}^{2}\right)$

Putting it all together:

$8 {x}^{6} - 27 {y}^{6}$

$= \left(\sqrt{2} x - \sqrt{3} y\right) \left(2 {x}^{2} + \sqrt{6} x y + 3 {y}^{2}\right) \left(\sqrt{2} x + \sqrt{3} y\right) \left(2 {x}^{2} - \sqrt{6} x y + 3 {y}^{2}\right)$

$\textcolor{w h i t e}{}$
Notes

What is interesting about this problem is that the coefficients $8 = {2}^{3}$ and $27 = {3}^{3}$ naturally point to treating this as a difference of cubes first to find:

$8 {x}^{6} - 27 {y}^{6} = \left(2 {x}^{2} - 3 {y}^{2}\right) \left(4 {x}^{4} + 6 {x}^{2} {y}^{2} + 9 {y}^{4}\right)$

If we then decide to allow irrational coefficients then the first of these factors fairly straightforwardly as:

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

but the second is not so straightforward...

$4 {x}^{4} + 6 {x}^{2} {y}^{2} + 9 {y}^{4}$

will not factor as a "quadratic in ${x}^{2}$ and ${y}^{2}$" with real coefficients.

To factor it, we can consider the following:

$\left({a}^{2} - k a b + {b}^{2}\right) \left({a}^{2} + k a b + {b}^{2}\right) = {a}^{4} + \left(2 - {k}^{2}\right) {a}^{2} {b}^{2} + {b}^{4}$

We can match ${a}^{4}$ with $4 {x}^{4}$ by putting $a = \sqrt{2} x$, ${b}^{4}$ with $9 {y}^{4}$ by putting $b = \sqrt{3} y$, to find:

$\left(2 {x}^{2} - \sqrt{6} k x y + 3 {y}^{2}\right) \left(2 {x}^{2} + \sqrt{6} k x y + 3 {y}^{2}\right)$

$= 4 {x}^{4} + \left(12 - 6 {k}^{2}\right) {x}^{2} {y}^{2} + 9 {y}^{4}$

So to match $12 - 6 {k}^{2}$ with $6$, we just need ${k}^{2} = 1$.

Using $k = 1$ we find:

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