<s>[INST] Simplifying $\frac{x^6-1}{x-1}$ I have this:
$$\frac{x^6-1}{x-1}$$
I know it can be simplified to $1 + x + x^2 + x^3 + x^4 + x^5$
Edit : I was wondering how to do this if I didn't know that it was the same as that.
 [/INST] Using complex numbers, you can factor $x^6-1$ from the roots of unit.
$$(x^6-1)=(x-1)(x-e^{i\pi/3})(x-e^{2i\pi/3})(x+1)(x-e^{4i\pi/3})(x-e^{5i\pi/3}).$$
Dropping the factor $(x-1)$ and grouping for the conjugate roots*,
$$(x^2-x+1)(x+1)(x^2+x+1)=(x^4+x^2+1)(x+1)=x^5+x^4+x^3+x^2+x+1.$$

*Using
$(x-z)(x-z^*)=x^2-(z+z^*)x+zz^*=x^2-2\Re(z)x+|z|^2.$
</s>