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.

You can easily see that $x^6 - 1$ has a root at $1$, so you know that
$$x^6 - 1 = (x-1) \cdot p(x)$$
Where $p(x)$ is a polynomial of degree $5$. Perform polynomial division to find $p$:
$$\begin{align*} (x^6 - 1) \div (x-1) & = x^5 + (x^5 - 1) \div (x-1) \\
& = x^5 + x^4 + (x^4 - 1) \div (x-1) \\
& = x^5 + x^4 + x^3 + (x^3 - 1) \div (x-1) \\
& = x^5 + x^4 + x^3 + x^2 + (x^2 - 1) \div (x-1) \\
& = x^5 + x^4 + x^3 + x^2 + x + (x-1) \div (x-1) \\
& = x^5 + x^4 + x^3 + x^2 + x + 1
\end{align*}$$
