A easier way to solve this equation $(z+1)^6+(z-1)^6=0$ in complex field?

Question

I use the binomial expansion and turn it into a cubic function to factorize it completely, but my teacher told me it could be much easier. Could you provide some other method to solve this equation or abbreviate my approach? $$(z-1)^6+(z+1)^6=0 \\z^6-6z^5+15z^4-20z^3+15z^2-6z+1+z^6+6z^5+15z^4+20z^3+15z^2+6z+1=0\\2z^6+30z^4+30z^2+2=0\\z^6+15z^4+15z^2+1=0\\z^4(z^2+1）+14z^2(z^2+1)+(z^2+1)=0\\(z^4+14z^2+1)(z^2+1)=0\\(z^2+7-4\sqrt3)(z^2+7+4\sqrt3)(z+i)(z-i)=0\\ [z-(\sqrt3-2)i][z+(\sqrt3-2)i](z+i)(z-i)[z+(\sqrt3+2)i][z-(\sqrt3+2)i]=0$$

Answer 1

Guide:

$$(z+1)^6 = -(z-1)^6$$

$$\left( \frac{z+1}{z-1}\right)^6=-1=\exp\left( (2k+1)\pi i\right)$$

$$\left( \frac{z+1}{z-1}\right)=-1=\exp\left( \frac{2k+1}6\pi i\right)$$

Answer 2

Take $\omega=\cos\left(\frac\pi6\right)+\sin\left(\frac\pi6\right)i=\frac{\sqrt3}2+\frac i2$ and notice that $\omega^6=-1$. Then\begin{align}(z+1)^6+(z-1)^6=0&\iff(z+1)^6=\bigl(\omega(z-1)\bigr)^6\\&\iff z+1=\omega(z-1)\vee z+1=\omega^2(z-1)\vee z+1=\omega^3(z-1)\vee\\&\phantom{\iff}\vee z+1=\omega^4(z-1)\vee z+1=\omega^5(z-1)\vee z+1=\omega^6(z-1).\end{align}

Answer 3

Your approach can be simplified. Since $a^6+b^6=(a^2+b^2)(a^4-a^2b^2+b^4)$, $(z-1)^6+(z+1)^6=2(1+z^2)((z+1)^4-(z+1)^2(z-1)^2+(z-1)^4)=2(1+z^2)(1+14z^2+z^4)$.
No need to factorize that complicated polynomial.

Answer 4

From $(z+1)^6+(z-1)^6=0$ we get $|z+1|=|z-1|$.

Observe that $|z+1|=|z-1| \iff z \in i \mathbb R$.

Can you proceed ?