<s>[INST] If $(x+1)^6=x^6$ then prove that $x=\frac{-1}2-i\cot (\frac{\theta}2)$ where $\theta=\frac{2k\pi}6$, $k=0,1,2,3,4,5$ QUESTION
If $(x+1)^6=x^6$ then prove that $x=\Large{\frac{-1}2}$$-i\cot (\large\frac\theta2)$ where $\theta=\Large\frac{2k\pi}6$, $k=0,1,2,3,4,5$

How can we prove this question using complex numbers. We need to prove it using roots of complex numbers.
If we consider$$\frac{(x+1)^{6}}{x^{6}} = 1$$
then $$\frac{(x+1)}{x}=(1)^\frac{1}{6} = \cos (\theta)^\frac{1}{6} + i \sin (\theta)^\frac{1}{6}$$
On further simplification I get the answer$$x=\frac{-1}{2}-\frac{i\cot(\theta/2)}{2}$$
But we need to prove $x=\frac{-1}{2}
-i\cot (\theta/2)$. What can be the error?
My solved example is here

 [/INST] First note that $k = 0$ does not give a solution, as $cot(0)$ is undefined. This should match with the intuition that your equation
$$(x+1)^6 = x^6 \Longrightarrow 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1 = 0$$
should only have up to 5 roots by the fundamental theorem of algebra.
Otherwise, your solution is sound. You can verify that the roots are
$$x_1 = -\frac{1}{2} - i \frac{\sqrt{3}}{2} = -\frac{1}{2} - \frac{i}{2} \cot(\frac{\pi}{6}),$$
$$x_2 = -\frac{1}{2} - i \frac{1}{2\sqrt{3}} = -\frac{1}{2} - \frac{i}{2} \cot(\frac{2\pi}{6}),$$
$$x_3 = -\frac{1}{2} = -\frac{1}{2} - \frac{i}{2} \cot(\frac{3\pi}{6}),$$
$$x_4 = -\frac{1}{2} + i\frac{1}{2\sqrt{3}} = -\frac{1}{2} - \frac{i}{2} \cot(\frac{4\pi}{6}),$$
$$x_5 = -\frac{1}{2} + i\frac{\sqrt{3}}{2} = -\frac{1}{2} - \frac{i}{2} \cot(\frac{5\pi}{6}).$$
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