<s>[INST] If $\alpha\ne1,\alpha^6=1$ and $\sum_{r=1}^6 {^6}C_r\alpha^{r-1}=x,$ then find the value of $|x|$. The following question is taken from the practice set of JEE exam.

If $\alpha\ne1,\alpha^6=1$ and $\sum_{r=1}^6 {^6}C_r\alpha^{r-1}=x,$ then find the value of $|x|$.

$$\sum_{r=1}^6 {^6}C_r\alpha^{r-1}=x\\\implies {^6}C_1\alpha^0+{^6}C_2\alpha^1+{^6}C_3\alpha^2+{^6}C_4\alpha^3+{^6}C_5\alpha^4+{^6}C_6\alpha^5=x\\\implies {^6}C_1\alpha^1+{^6}C_2\alpha^2+{^6}C_3\alpha^3+{^6}C_4\alpha^4+{^6}C_5\alpha^5+{^6}C_6\alpha^6=\alpha x\\\implies {^6}C_0\alpha^0+{^6}C_1\alpha^1+{^6}C_2\alpha^2+{^6}C_3\alpha^3+{^6}C_4\alpha^4+{^6}C_5\alpha^5+{^6}C_6\alpha^6=\alpha x+{^6}C_0\\\implies (1+\alpha)^6=1+\alpha x$$
How shall I proceed next? I understand $\alpha$ is $6^{th}$ root of unity. How to use that here? Maybe $\alpha^6=1$ can be written as $$\alpha^6-1=0\\\implies (\alpha-1)(\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1)=0$$
Also, $\alpha\ne1\implies \alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1=0$. Is this of some help here? Thanks.
 [/INST] I'll do the case where $\alpha$ is a primitive $6$-th root of unity.

Then $\alpha$ satisfies $\alpha^2-\alpha+1=0$,$\;$hence
\begin{align*}
x&=\sum_{i=1}^6 {\small{\binom{6}{i}}}\alpha^{i-1}
\\[4pt]
&=
\alpha^5+6\alpha^4+15\alpha^3+20\alpha^2+15\alpha+6
\\[4pt]
&=
(\alpha^3+7\alpha^2+21\alpha+34)(\alpha^2-\alpha+1)+(28\alpha-28)
\\[4pt]
&=
28\alpha-28
&&
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!
\bigl(\text{since $\alpha^2-\alpha+1=0$}\bigr)
\\[4pt]
&=
28(\alpha-1)
\\[4pt]
&=
28\alpha^2
&&
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!
\bigl(\text{$\alpha^2-\alpha+1=0$ implies $\alpha-1=\alpha^2$}\bigr)
\\[10pt]
\implies\;|x|&=
|28\alpha^2|
\\[4pt]
&=
28|\alpha|^2
\\[4pt]
&=
28
&&
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!
\bigl(\text{since $|\alpha|=1$}\bigr)
\end{align*}
</s>