Let $$M=\begin{pmatrix} 0 & 1 & 1 & 1\\ 1 & 0 & \alpha + \beta & \alpha + \gamma\\ 1 & \beta + \alpha & 0 & \beta + \gamma\\ 1& \gamma + \alpha & \gamma + \beta & 0 \end{pmatrix},$$ then it holds $$\det(M) = −4(\alpha\beta + \beta\gamma + \gamma\alpha).$$
What is the value of this when $\alpha,\beta,\gamma$ are the three roots of the equation $x^3 − 1 = 0$?
Can anyone help me to do it by elementary row operation?
My idea is just solving $\alpha,\beta,\gamma$ and then plug in $−4(\alpha\beta + \beta\gamma + \gamma\alpha)$ and when I try to find the determinant of the LHS I got $-4\beta\gamma$.