Question

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

Answer 1

`8x^6-27y^6=2^3(x^2)^3-3^3(y^2)^3=(2x^2)^3-(3y^2)^3=(2x^2-3y^2)(4x^4+6x^2y^2+9y^4)`

Explanation:

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

`8x^6-27y^6=2^3(x^2)^3-3^3(y^2)^3=(2x^2)^3-(3y^2)^3`

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

`A^3-B^3=(A-B)(A^2+AB+B^2)`

so let's do that.

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

Answer 2

`8x^6-27y^6`

`= (sqrt(2)x-sqrt(3)y)(2x^2+sqrt(6)xy+3y^2)(sqrt(2)x+sqrt(3)y)(2x^2-sqrt(6)xy+3y^2)`

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 = (a-b)(a+b)`

Difference of cubes:

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

Sum of cubes:

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

Using the difference of squares:

`8x^6-27y^6 = (2sqrt(2)x^3)^2-(3sqrt(3)y^3)^2`

`color(white)(8x^6-27y^6) = (2sqrt(2)x^3-3sqrt(3)y^3)(2sqrt(2)x^3+3sqrt(3)y^3)`

`color(white)(8x^6-27y^6) = ((sqrt(2)x)^3-(sqrt(3)y)^3)((sqrt(2)x)^3+(sqrt(3)y)^3)`

Using the difference of cubes:

`(sqrt(2)x)^3-(sqrt(3)y)^3 = (sqrt(2)x-sqrt(3)y)((sqrt(2)x)^2+(sqrt(2)x)(sqrt(3)y)+(sqrt(3)(y))^2)`

`color(white)((sqrt(2)x)^3-(sqrt(3)y)^3) = (sqrt(2)x-sqrt(3)y)(2x^2+sqrt(6)xy+3y^2)`

Using the sum of cubes:

`(sqrt(2)x)^3+(sqrt(3)y)^3 = (sqrt(2)x+sqrt(3)y)((sqrt(2)x)^2-(sqrt(2)x)(sqrt(3)y)+(sqrt(3)(y))^2)`

`color(white)((sqrt(2)x)^3+(sqrt(3)y)^3) = (sqrt(2)x+sqrt(3)y)(2x^2-sqrt(6)xy+3y^2)`

Putting it all together:

`8x^6-27y^6`

`= (sqrt(2)x-sqrt(3)y)(2x^2+sqrt(6)xy+3y^2)(sqrt(2)x+sqrt(3)y)(2x^2-sqrt(6)xy+3y^2)`

`color(white)()`
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:

`8x^6-27y^6 = (2x^2-3y^2)(4x^4+6x^2y^2+9y^4)`

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

`2x^2-3y^2 = (sqrt(2)x-sqrt(3)y)(sqrt(2)x+sqrt(3)y)`

but the second is not so straightforward...

`4x^4+6x^2y^2+9y^4`

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

To factor it, we can consider the following:

`(a^2-kab+b^2)(a^2+kab+b^2) = a^4+(2-k^2)a^2b^2+b^4`

We can match `a^4` with `4x^4` by putting `a=sqrt(2)x`, `b^4` with `9y^4` by putting `b=sqrt(3)y`, to find:

`(2x^2-sqrt(6)kxy+3y^2)(2x^2+sqrt(6)kxy+3y^2)`

`= 4x^4+(12-6k^2)x^2y^2+9y^4`

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

Using `k=1` we find:

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