首先看方程:
z^7-1=0

除了z=1外,其餘幾個解是互為倒數的共軛複數,其實數部分就是以上三個數。作分解得方程

z^6+z^5+z^4+z^3+z^2+z+1=0,  since z=0 is not the solution, divide it by z^3, we have

z^3+z^2+z+1+1/z+1/z^2+1/z^3=0.

Let t=z+1/z, we get finally t^3+t^2-2t-1=0.

since z=(1/z)*(conjugation), t has 3 solutions which are double of the above 3 values. Then 

a=-1/2, b=-2/4=-1/2, c=1/8, abc=1/32.

This avoids complecated calculation of trigonometry functions.