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  4. The value of x obtained from the equation |x+α&β&γ γ&x+β&α α&β&x+γ|=0 will be

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Q. The value of $x$ obtained from the equation $\begin{vmatrix}x+\alpha&\beta&\gamma\\ \gamma&x+\beta&\alpha\\ \alpha&\beta&x+\gamma\end{vmatrix}=0$ will be

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Solution:

Given $\begin{vmatrix}x+\alpha&\beta&\gamma\\ \gamma &x+\beta&\alpha\\ \alpha&\beta& x+\gamma\end{vmatrix}=0$
Operate $C_{1} \to C_{1}+C_{2}+C_{3}$
$\begin{vmatrix}x+\alpha+\beta+\gamma&\beta&\gamma\\ x+\alpha+\beta+\gamma &x+\beta&\alpha\\ x+\alpha+\beta+\gamma&\beta &x+\gamma\end{vmatrix}=0$
$=\left(x+\alpha+\beta+\gamma\right) \begin{vmatrix}1&\beta&\gamma\\ 1&x+\beta&\alpha\\ 1&\beta &x+\gamma\end{vmatrix}=0$
$\Rightarrow x+\alpha+\beta+\gamma=0 \Rightarrow \, x=-\left(\alpha+\beta+\gamma\right)$ Again if
$\begin{vmatrix}1&\beta&\gamma\\ 1&x+\beta&\alpha\\ 1&\beta&\gamma\end{vmatrix}=0 \Rightarrow \begin{vmatrix}1&\beta&\gamma\\ 0&x&\alpha-\gamma\\ 0&0&x\end{vmatrix}=0$
$\Rightarrow x^{2}=0 \Rightarrow x=0$
$\therefore $ Solutions of the equation are $x = 0$,
$-\left(\alpha+\beta+\gamma\right)$