1. ## Factorisation

Hi.

How would go about factoring $\displaystyle x^6+1$

Thanks

2. what about $\displaystyle (x^3+i)(x^3-i)$

expanding back out to check

$\displaystyle x^6+x^3i-x^3i-i^2$

$\displaystyle x^6-i^2$

$\displaystyle x^6-(-1)$

$\displaystyle x^6+1$

3. Originally Posted by deltaxray
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...

4. You can factorise it to:

$\displaystyle ( x^2+\sqrt{3}x+1 ) ( x^2-\sqrt{3}x+1 ) ( x^2+1 )$

But unfortunately I cant remember the rule for doing this

5. Originally Posted by Stroodle
You can factorise it to:

$\displaystyle ( x^2+\sqrt{3}x+1 ) ( x^2-\sqrt{3}x+1 ) ( x+1 )$

But unfortunately I cant remember the rule for doing this
You have a typo. The (x + 1) should be (x^2 + 1). See below.

Originally Posted by deltaxray
Hi.

How would go about factoring $\displaystyle x^6+1$

Thanks
$\displaystyle x^6 + 1 = (x^2)^3 + 1^3 = (x^2 + 1)((x^2)^2 - x^2 + 1^2)$

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.