Find all five solutions of the equation $z^5+z^4+z^3+z^2+z+1 = 0$ $z^5+z^4+z^3+z^2+z+1 = 0$
I can't figure this out can someone offer any suggestions?
Factoring it into $(z+1)(z^4+z^2+1)$ didn't do anything but show -1 is one solution. 
I solved for all roots of $z^4 = -4$ but the structure for this example was more simple.

Here is another way to pursue this (just for fun!).
Recall that for $|x| < 1$ we have:
$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$$
Multiplying both sides by $x^6$ yields:
$$\frac{x^6}{1-x} = x^6 + x^7 + x^8 + x^9 + \cdots$$
Now we subtract the former line from the latter:
$$\frac{x^6 - 1}{1-x} = -(1 + x + x^2 + x^3 + x^4 + x^5)$$
We now multiply both sides by $-1$ to obtain your original expression:
$$\frac{x^6 - 1}{x-1} = 1 + x + x^2 + x^3 + x^4 + x^5$$
In one way, this is no different than the suggestion in your accepted answer. But I think (hope) seeing the connection arise in a different manner is of interest. Anyway, at this point you seek to find the roots of $x^6 - 1 = (x^3 + 1)(x^3 - 1)$, and you can now factor the latter two parenthetical expressions as the sum and difference of cubes, respectively, to find the roots. 
(Noting that/why $x\neq 1$.)
