Question

Factoring the sum or difference of two cubes

But what if the exponents are not cubes...
Example x^6 - y^6

Answer 1

Well the nice this about this one is that if you first use difference of two squares, then
\[
x^6-y^6=(x^3)^2-(y^3)^2 = (x^3-y^3)(x^3+y^3)
\]and lo and behold, you have a difference and a sum of two cubes.

Answer 2

Actually, they are cubes. \(x^6=\left(x^2\right)^3\).

Answer 3

Ahh ok, so for 27x^6 + 64x^12, the answer would be (3x^3 - 4^6) (3x^3 + 4^6)?

Answer 4

Not exactly. If you want to factor \(27x^6 + 64x^{12}\), the best this to do it to factor out \(x^6\), which gives you \(x^6(27 + 64x^{6})\), and now what's inside the bracket is a sum of two cubes.

Answer 5

*the best thing to do is...

Answer 6

Would my answer still be correct?

Answer 7

The answer (3x^3 - 4^6) (3x^3 + 4^6)? No - you can check this yourself. If you multiply it out you don't get 27x^6 + 64x^12.

Answer 8

Ok, so how do I know whether to use the coefficient or the exponent when factoring?

Answer 9

Well, the first thing to do when you're factoring is to take out any obvious common factors, such as \(x^6\) in \(27x^6 + 64x^{12}\). This then left you with a sum of two cubes, which you can tackle.

With \(x^6 - y^6\), there are no obvious common factors. You can then proceed in two ways. Either notice that is is the difference of two squares, as I did, or you can in fact notice that it is the difference of two cubes, so it is
\[x^6 - y^6 = (x^2)^3-(y^2)^3 = (x^2-y^2)(x^4+x^2y^2+y^4).
\]I prefer the first method, because \((x^4+x^2y^2+y^4)\) isn't so easy to factor.

Answer 10

Ok, thank you so much!