Question

How do you factor `1331s^12 - 1728t^6`?

Answer 1

`1331s^12 - 1728t^6`

`=(sqrt(11)s^2-2sqrt(3)t)(11s^4+2sqrt(33)s^2t+12t^2)(sqrt(11)s^2+2sqrt(3)t)(11s^4-2sqrt(33)s^2t+12t^2)`

Explanation:

We will 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)`

First let us find out how to factorise `a^6-b^6`

We can treat this as a difference of squares, then difference and sum of cubes as follows:

`a^6-b^6`

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

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

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

So we have our very own difference of sixth powers identity:

`a^6-b^6 = (a-b)(a^2+ab+b^2)(a+b)(a^2-ab+b^2)`

Note that:

• `1331 = 11^3 = (sqrt(11))^6`

• `1728 = 12^3 = (2^2*3)^3 = (2sqrt(3))^6`

Let's put our difference of sixth powers to work with `a = sqrt(11)s^2` and `b=2sqrt(3)t`

`a^6 - b^6`

`= (sqrt(11)s^2)^6 - (2sqrt(3)t)^6`

`= 11^3s^12 - 12^3t^6`

`= 1331s^12 - 1728t^6`

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

`=(sqrt(11)s^2-2sqrt(3)t)(11s^4+2sqrt(33)s^2t+12t^2)(sqrt(11)s^2+2sqrt(3)t)(11s^4-2sqrt(33)s^2t+12t^2)`