Show that $\sqrt{-1-\sqrt{-1-{\sqrt{-1...}}}}$ can be expressed in the form $\alpha$ or $\alpha^2$, Hence prove $$(a+b\alpha+c\alpha^2)(a+b\alpha^2+c\alpha)=\frac12[(a-b)^2+(b-c)^2+(c-a)^2]$$
My try:
I wrote the expression as,$$y=\sqrt{-1-\sqrt{-1-{\sqrt{-1...}}}}$$ $$y^2+y+1=0$$ $$y=\frac{-1+\sqrt{3}i}{2}\text{ or } y=\frac{-1-\sqrt{3}i}{2}$$ This is in the form of $\alpha$ and $\alpha^2.$I have trouble solving the second part. Can someone please explain it to me? Thank you in advance.
The LHS is $a^2+b^2 \alpha^3+c^2 \alpha^3+ab(\alpha^2+\alpha)+ac(\alpha^2+\alpha)+bc(\alpha^4+\alpha)$.
Use the fact that $\alpha^4=\alpha$,$\alpha^2+\alpha=-1$,$\alpha^3=1$.
So The LHS becomes $$a^2+b^2+c^2-ab-ac-bc\\=\frac 1 2\left[a^2+b^2-2ab+b^2+c^2-2bc+a^2+c^2-2ac\right]\\=\frac 1 2\left[(a-b)^2+(b-c)^2+(a-c)^2\right]$$