Solve the equations

$\left|\begin{array}{ccc}x+a & x & x \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0$

Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}$, we get:

$\left|\begin{array}{ccc}3 x+a & 3 x+a & 3 x+a \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0$

$\Rightarrow(3 x+a)\left|\begin{array}{ccc}1 & 1 & 1 \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0$

Applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}$ and $\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}$, we have:

$(3 x+a)\left|\begin{array}{lll}1 & 0 & 0 \\ x & a & 0 \\ x & 0 & a\end{array}\right|=0$

Expanding along $R_{1}$, we have:

$(3 x+a)\left[1 \times a^{2}\right]=0$

$\Rightarrow a^{2}(3 x+a)=0$

But $a \neq 0$.

Therefore, we have:

$3 x+a=0$

$\Rightarrow x=-\frac{a}{3}$