What is the prime factor of x^6 + 1 ?

Question

hitch · Answer

$$x ^{6}+1 = (x ^{2}+x+1)(x ^{2}−x+1)(x+1)(x−1)$$

zepdrix · Answer

Hey Dreamie :)
We can make use of our `Sum of Cubes Identity`:
$\large\rm a^3+b^3=(a+b)(a^2-ab+b^2)$

We need to rewrite this 6th power as a cube though.
We can do this little trick using rules of exponents,$$\large\rm x^6=(x^2)^3$$Ahh that should help!
So our expression,$$\large\rm x^6+1$$is really,$$\large\rm (x^2)^3+1^3$$and we can apply our identity, yay!$$\large\rm =(x^2+1)\left((x^2)^2-(x^2)(1)+1^2\right)$$

zepdrix · Answer

So we have,$$\large\rm (x^2+1)(x^4-x^2+1)$$Which uhhh, ya I think both of those are prime factors.

I'm not quite sure how Hitch factored it down so far >.< hmm

hitch · Answer

it is true by the end Mr.Zepdrix

dreamie98 · Answer

thanks for the help guys, I couldn't answer because my internet went out D: <3