Problem: Find the roots of $6z^5+15z^4+20z^3+15z^2+6z+1 = 0$.
What I found: I realized that the coefficients were the binomial coefficients of $6$. Putting these values in, you would get $\dbinom{6}{1}z^5 + \dbinom{6}{2}z^4 + \dbinom{6}{3}z^3 + \dbinom{6}{4}z^2 + \dbinom{6}{5}z + \dbinom{6}{6} = 0$.
I also thought that maybe there was some applications of the roots of unity here, but I wasn't quite sure. Some help here would be really awesome. Thanks!
Hint : Rewrite your equation as $(z+1)^6-z^6=0$ and since $z=0$ isn't a solution, show that $\frac{z+1}{z}$ is a 6-th root of unity.