x^6+1=(x^4-x^2+1)(x^2+1)
How did I find this? First of all, substitute x^2 = y to get y^3 + 1.
Notice that y = -1 is a solution of y^3+1 = 0, so (y+1) is a factor.
We can easily derive y^3+1=(y^2-y+1)(y+1), noticing the way that the intermediate terms cancel out when you multiply this out.
Substituting y = x^2 back in, we get:
x^6+1=(x^4-x^2+1)(x^2+1)
How do we know this factorization is complete?
There is no linear factor, because x^6+1 is always > 0 (for real values of x).
How about quadratic factors?
Suppose x^4-x^2+1=(x^2+ax+b)(x^2+cx+d).
Looking at the coefficient of x^3, we get a+c = 0, so c = -a:
... =(x^2+ax+b)(x^2-ax+d)
The constant term gives us that bd = 1 so b = d = 1 or b = d = -1.
The coefficient of x^2 gives us a^2-(b+d)=1. Hence a^2 = 3 or a^2 = -1. Neither have integer solutions.
