# Factorisation

• Nov 29th 2009, 11:52 PM
deltaxray
Factorisation
Hi.

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

Thanks
• Nov 30th 2009, 12:45 AM
pickslides
what about $\displaystyle (x^3+i)(x^3-i)$

expanding back out to check

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

$\displaystyle x^6-i^2$

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

$\displaystyle x^6+1$
• Nov 30th 2009, 12:51 AM
Prove It
Quote:

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...
• Nov 30th 2009, 12:57 AM
Stroodle
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 :(
• Nov 30th 2009, 03:09 AM
mr fantastic
Quote:

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.

Quote:

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.