The value of x obtained from the equation $\left|\begin{array}{ccc}x+\alpha & \beta & \gamma \\ \gamma & x+\beta & \alpha \\ \alpha & \beta & x+\gamma \end{array}\right|=0$ will be

Given $\left|\begin{array}{ccc}x+\alpha & \beta & \gamma \\ \gamma & x+\beta & \alpha \\ \alpha & \beta & x+\gamma \end{array}\right|=0$

Operate ${\mathrm{C}}_1\to {\mathrm{C}}_1+{\mathrm{C}}_2+{\mathrm{C}}_3$

$\begin{array}{r}\left|\begin{array}{ccc}x+\alpha +\beta +\gamma & \beta & \gamma \\ x+\alpha +\beta +\gamma & x+\beta & \alpha \\ x+\alpha +\beta +\gamma & \beta & x+\gamma \end{array}\right|=0\\ =(x+\alpha +\beta +\gamma )\left|\begin{array}{ccc}1& \beta & \gamma \\ 1& x+\beta & \alpha \\ 1& \beta & x+\gamma \end{array}\right|=0\\ \Rightarrow x+\alpha +\beta +\gamma =0\Rightarrow x=-(\alpha +\beta +\gamma )\end{array}$

Again if 

$\begin{array}{l}\left|\begin{array}{ccc}1& \beta & \gamma \\ 1& x+\beta & \alpha \\ 1& \beta & x+\gamma \end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}1& \beta & \gamma \\ 0& x& \alpha -\gamma \\ 0& 0& x\end{array}\right|=0\\ \Rightarrow x^2=0\Rightarrow x=0\end{array}$

$\therefore$ Solutions of the equation are $x=0,-(\alpha +\beta +\gamma )$