Hi.
How would go about factoring $\displaystyle x^6+1$
Thanks
If you're factorising over the Real Numbers, it does not factorise.
If you're factorising over the Complex Numbers, change it to a difference of two squares.
$\displaystyle x^6 + 1 = (x^3)^2 - i^2$
$\displaystyle = (x^3 + i)(x^3 - i)$.
You may be able to use a sum or difference of two cubes factorisation now as well...
You have a typo. The (x + 1) should be (x^2 + 1). See below.
$\displaystyle x^6 + 1 = (x^2)^3 + 1^3 = (x^2 + 1)((x^2)^2 - x^2 + 1^2)$
using the sum of two cubes formula
$\displaystyle = (x^2 + 1)(x^4 - x^2 + 1)$.
Now note that $\displaystyle x^4 - x^2 + 1 = x^4 + 2x^2 + 1 - 3x^2 = (x^2 + 1)^2 - 3x^2 = (x^2 + 1 - \sqrt{3} x) (x^2 + 1 + \sqrt{3} x)$ using the difference of two squares formula.