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.

Hint:- $$y^3-1=y^2(y-1)+y(y-1)+(y-1)=(y-1)\left(y^2+y+1\right)$$
Solution:-

$y=x^2\implies x^6-1=\left(x^2-1\right)\left(x^4+x^2+1\right)=(x-1)(x+1)\left(x^4+x^2+1\right)$ $$\boxed{\therefore\dfrac{x^6-1}{x-1}=(x+1)\left(x^4+x^2+1\right)=x^5+x^4+x^3+x^2+x+1}$$

