[tex]\alpha^2+\alpha +1 = 0\Big|\cdot(\alpha -1),\quad \alpha \neq 1 \\ \\\Rightarrow (\alpha -1)(\alpha^2+\alpha +1) = 0\\ \\ \Rightarrow \alpha^3-1 = 0\\ \\ \Rightarrow \alpha^3 = 1\\ \\ \\ \big[(1+\alpha)^{6n+2}+1\big]^2-\alpha= \left\{\Big[(1+\alpha)^{2}\Big]^{3n}\cdot(1+\alpha)^2+1\right\}^2-\alpha =\\ \\=\left[(\alpha^2+2\alpha+1)^{3n}\cdot(\alpha^2+2\alpha+1)+1\right]^2-\alpha=\\\\=\left(\alpha^{3n}\cdot\alpha+1\right)^2-\alpha =[/tex]
[tex]=\Big[(\alpha^3)^n \cdot \alpha+1\Big]^2-\alpha=(1\cdot \alpha+1)^2-\alpha =\\ \\= \alpha^2+2\alpha+1-\alpha =\alpha^2+\alpha+1+\alpha - \alpha = \\ \\ = 0+\alpha - \alpha = \boxed{0}[/tex]
