Ok so I need to solve;

 

[math]

 

z^{5} + x^{4} + z^{3}+z^{2}+z + 1 = 0

 

[/math]

 

My first thought is to use the summation equation where a = 1, i.e.;

 

[math]

 

\frac {1-z^5}{1-z} = 0

 

[/math]

 

but then I'm not sure where to go from there; in the past I have had something^something else = 1 then something = alpha and then made z = f(alpha) however here there isnt a whole term to the same index.

 

Any ideas?

My first idea is that three times an odd power of z plus two times an even power of z (I assume x^4 was a typo) shall equal -1 ...

First thing: That's [math]\frac{1-z^6}{1-z}=0[/math], not [math]\frac{1-z^5}{1-z}=0[/math].

 

Multiple both sides by the denominator. [math]1-x^6=0[/math] has six solutions, one of which results in 0/0. The remaining five solutions are the five solutions to the original problem.

Yep, that checks out.

 

Cheers for that.