How to factorise $x^6 + 1$? How do I factorise this? I already said that
$$x^6 + 1 = 0 \implies (x^3)^2 + 1 = 0$$
so we then get
$$(x^3)^2 = - 1$$
and then there are no real roots. Similar thing happens if I try it as $(x^2)^3$.
How would I then go about factorising this? I know its possible as 
$$(x^6 + 1) = (x^2 + 1)(x^4 - x^2 + 1).$$

If $x^6+1=0$, then $x^3$ must be either $i$ or $-i$ (because $(x^3)^2=x^6=-1$). Hence, one must find the cube roots of $\pm i$. The numbers $i$ and $-i$ are easily seen to be cube roots of each other.
Now, $(a+bi)^3=a^3+3a^2bi+3a(bi)^2+(bi)^3=a^3+3a^2bi+3ab^2i^2+b^3i^3=a^3+3a^2bi-3ab^2-b^3i=(a^3-3ab^2)+(3a^2b-b^3)i$. If this were to equal $i$, then the real part $a^3-3ab^2$ would have to be zero. Hence, either $a=0$ or $a^2-3b^2=0$. If $a=0$, then one gets the solution $-i$. For the other two solutions, substitute $a=\pm b\sqrt{3}$ into the equation $3a^2b-b^3=1$ to get $3(3b^2)b-b^3=1$. Simplifying the left side, one gets $8b^3=1$. The only real solution for $8b^3=1$ is $b=\frac{1}{2}$.
Hence, the other two cube roots of $i$ (besides $-i$) are $\frac{1}{2}\sqrt{3}+\frac{1}{2}i$ and $-\frac{1}{2}\sqrt{3}+\frac{1}{2}i$. Conjugating those two cube roots of $i$, one gets $\frac{1}{2}\sqrt{3}-\frac{1}{2}i$ and $-\frac{1}{2}\sqrt{3}-\frac{1}{2}i$ as the two cube roots of $-i$ besides $i$.
The fully factorized form of $x^6+1$ (over the complex numbers) is therefore $(x-i)(x+i)(x-\frac{1}{2}\sqrt{3}-\frac{1}{2}i)(x+\frac{1}{2}\sqrt{3}-\frac{1}{2}i)(x-\frac{1}{2}\sqrt{3}+\frac{1}{2}i)(x+\frac{1}{2}\sqrt{3}+\frac{1}{2}i)$.
