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.
 A: You have correctly derived
$$ (1+\alpha)^6 = 1+\alpha x $$
(though it would have been easier to follow that derivation if you had kept the indexed sum notation in the intermediate lines).
From here it is probably simplest to leave the algebra and polynomials aside and simply calculate the value of the left-hand side based on which of the sixth roots of $1$ you have as $\alpha$. Using geometry to express $1+\alpha$ in polar coordinates in each case:
$$ \begin{array}{ccc} \alpha & 1 + \alpha & (1+\alpha)^6 \\ \hline
e^{i\pi/3} & \sqrt3 e^{i\pi/6} & -27 \\
e^{2i\pi/3} & e^{i\pi/3} & 1 \\
e^{3i\pi/3}=-1 & 0 & 0 \end{array} $$
and the last two roots are complex conjugates of the first.
Since $|\alpha|=1$ we have $|x|=|\alpha x|=|(1-\alpha)^6-1|$ which you can now compute directly.
As TonyK noted, the result depends on which root you choose for $\alpha$, so the question is strictly speaking ill-formed.
A: $\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1=0$
$\implies\alpha^3(\alpha^2+\alpha+1)+(\alpha^2+\alpha+1)=0$
$\implies(\alpha^3+1)(\alpha^2+\alpha+1)=0$
$\implies\alpha^3=-1, \alpha^2+\alpha+1=0$
Now from your equation, $(1+\alpha)^6=1+x\alpha$
$(1+\alpha)^6-1=x\alpha$
$\alpha(\alpha+2)(\alpha^2+3\alpha+3)(\alpha^2+\alpha+1)=x\alpha$
$0=x\alpha$
$\therefore|x|=0$
Here I didn't talked about the equation $\alpha^3=-1$. So this is just a special case.
A: 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*}
A: @Ranjit Kumar Sarkar gave the hint that $$\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1=\alpha^3(\alpha^2+\alpha+1)+\alpha^2+\alpha+1=(\alpha^3+1)(\alpha^2+\alpha+1)$$
Equating that to zero, I get $$\alpha=\omega,\omega^2,-1,-\omega,-\omega^2$$
where $\omega$ and $\omega^2$ are the complex cube roots of unity.
Also, I had already got that $$(1+\alpha)^6=1+\alpha x ---(1)$$
@Troposphere suggested that I would get this equation faster if I had kept it in summation form. So, multiplying the given equation by $\alpha$, I get $$\sum_{r=1}^6 {^6}C_r\alpha^{r}=\alpha x\\\implies\sum_{r=0}^6 {^6}C_r\alpha^{r}=1+\alpha x\\\implies(1+\alpha)^6=1+\alpha x$$
Putting $\alpha=\omega$ in $(1)$, I get $$(1+\omega)^6=1+\omega x\\\implies (-\omega^2)^6=1+\omega x\\\implies 1=1+\omega x\\\implies|x|=0$$
I get the same result for $\alpha=\omega^2$
If I put $\alpha=-1$ in $(1)$, I get $$0=1-x\implies|x|=1$$
If I put $\alpha=-\omega$ in $(1)$, I get $$(1-w)^6=1-\omega x\\\implies\left(1-\frac{-1+\sqrt3i}2\right)^6=1-\omega x\\\implies\left(\frac{3-\sqrt3i}2\right)^6=1-\omega x\\\implies(\sqrt3)^6\left(\frac{1+\sqrt3i}{2i}\right)^6=1-\omega x\\\implies\frac{(\sqrt3)^6}{i^6}(-\omega^2)^6=1-\omega x\\\implies-27=1-\omega x\\\implies\omega x=28\implies|x|=28$$
I get the same result for $\alpha=-\omega^2$
Thankyou @Ranjit Kumar Sarkar, @quasi, @TonyK and @Troposphere for helping me think through it. Thanks.
