If $\alpha$ and $\beta$ are non real numbers satisfying $x^3 -1 = 0$ , then evaluate, \begin{vmatrix} \lambda +1 &\alpha &\beta\\ \alpha &\lambda + \beta &1\\ \beta &1 &\lambda + \alpha \end{vmatrix}
I tried this: $\alpha $ =$\omega$, $\beta = \omega^2$. Then I modified columns as $C_1 = C_1+C_2+C_3$ and further $R_1=R_1-R_3$ and $R_2=R_2-R_3$. to get this: \begin{vmatrix} 0 &\omega-1 &\omega ^2-\lambda-\omega\\ 0 & \omega^2+\lambda+1 &\lambda+\omega+1\\ \omega &1 &\lambda + \omega \end{vmatrix} I further simplified it to get the value of determinant as $\lambda[\lambda^2-1-\omega]$ .The answer is only $\lambda^3.$ Please point out any errors if you find them.
Using your notation but with $\;t=\lambda\;$ :$${}$$
$$\begin{vmatrix} t + 1&\alpha &\beta\\ \alpha &t + \beta &1\\ \beta &1 &t + \alpha \end{vmatrix}=\begin{vmatrix} t + 1&w &w^2\\ w &t +w^2 &1\\ w^2 &1 &t +w \end{vmatrix}=$$$${}$$
$$=(t+1)(t^2-t+1)+1+1-(t+1)-w^2(t+w)-w(t+w^2)=$$
$$=t^3+1+2-t-1-w^2t-1-wt-1=t^3-(w^2+w+1)t=t^3$$
Observe that $\;\alpha+\beta+1=w+w^2+1=0\;$ and from here the above relations.