Using properties of determinants, solve the following for x.

$\left|\begin{array}{ccc}\mathrm{x}-2& 2\mathrm{x}-3& 3\mathrm{x}-4\\ \mathrm{x}-4& 2\mathrm{x}-9& 3\mathrm{x}-16\\ \mathrm{x}-8& 2\mathrm{x}-27& 3\mathrm{x}-64\end{array}\right|=0.$

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Using properties of determinants, solve the following for x.

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

First, apply some properties, so that when we expand the determinant, it is easy to simplify.
Given,  

$\left|\begin{array}{ccc}\mathrm{x}-2& 2\mathrm{x}-3& 3\mathrm{x}-4\\ \mathrm{x}-4& 2\mathrm{x}-9& 3\mathrm{x}-16\\ \mathrm{x}-8& 2\mathrm{x}-27& 3\mathrm{x}-64\end{array}\right|=0.$

On applying ${\mathrm{R}}_1 \to {\mathrm{R}}_1-{\mathrm{R}}_2 \mathrm{and} {\mathrm{R}}_2\to {\mathrm{R}}_2-{\mathrm{R}}_3,$ we get

$\Rightarrow \left|\begin{array}{ccc}2& 6& 12\\ 4& 18& 48\\ \mathrm{x}-8& 2\mathrm{x}-27& 3\mathrm{x}-64\end{array}\right|=0$

On taking 2 common from ${\mathrm{R}}_1$ and  ${\mathrm{R}}_2$, we get

$\Rightarrow 2\times 2\left|\begin{array}{ccc}1& 3& 6\\ 2& 9& 24\\ \mathrm{x}-8& 2\mathrm{x}-27& 3\mathrm{x}-64\end{array}\right|=0$

on applying ${\mathrm{R}}_2\to {\mathrm{R}}_2-2{\mathrm{R}}_1$ , we get

$\Rightarrow 4\left|\begin{array}{ccc}1& 3& 6\\ 0& 3& 12\\ \mathrm{x}-8& 2\mathrm{x}-27& 3\mathrm{x}-64\end{array}\right|=0$

On expanding along first column , we get

$$\begin{gathered} \Rightarrow 4\left[3(3\mathrm{x}-64)-12(2\mathrm{x}-27)+(\mathrm{x}-8)(3\times 12-3\times 6)\right]=0 \\ \Rightarrow 4\left[9\mathrm{x}-192-24\mathrm{x}+324+(\mathrm{x}-8)18\right]=0 \\ \Rightarrow =(3\mathrm{x}-12)=0 \\ \therefore \mathrm{x}=4 \end{gathered}$$

$$\begin{gathered} \mathrm{Given}, \left|\begin{array}{ccc}\mathrm{x}+\mathrm{a}& \mathrm{x}& \mathrm{x}\\ \mathrm{x}& \mathrm{x}+\mathrm{a}& \mathrm{x}\\ \mathrm{x}& \mathrm{x}& \mathrm{x}+\mathrm{a}\end{array}\right|=0 \\ \mathrm{On} \mathrm{applying} {\mathrm{C}}_1\to {\mathrm{C}}_1+{\mathrm{C}}_2+{\mathrm{C}}_3, \mathrm{we} \mathrm{get} \\ \left|\begin{array}{ccc}3\mathrm{x}+\mathrm{a}& \mathrm{x}& \mathrm{x}\\ 3\mathrm{x}+\mathrm{a}& \mathrm{x}+\mathrm{a}& \mathrm{x}\\ 3\mathrm{x}+\mathrm{a}& \mathrm{x}& \mathrm{x}+\mathrm{a}\end{array}\right|=0 \\ \mathrm{On} \mathrm{taking} (3\mathrm{x}+\mathrm{a}) \mathrm{common} \mathrm{from} {\mathrm{C}}_1, \mathrm{we} \mathrm{get} \\ (3\mathrm{x}+\mathrm{a})\left|\begin{array}{ccc}1& \mathrm{x}& \mathrm{x}\\ 1& \mathrm{x}+\mathrm{a}& \mathrm{x}\\ 1& \mathrm{x}& \mathrm{x}+\mathrm{a}\end{array}\right| =0 \\ \mathrm{Now}, \mathrm{on} \mathrm{applying} {\mathrm{R}}_2\to {\mathrm{R}}_2-{\mathrm{R}}_1 \mathrm{and} {\mathrm{R}}_3\to {\mathrm{R}}_3-{\mathrm{R}}_1, \mathrm{we} \mathrm{get} \\ (3\mathrm{x}+\mathrm{a})\left|\begin{array}{ccc}1& \mathrm{x}& \mathrm{x}\\ 0& \mathrm{a}& 0\\ 0& 0& \mathrm{a}\end{array}\right|=0 \end{gathered}$$

On expanding along ${\mathrm{C}}_1$, we get

$$\begin{gathered} (3\mathrm{x}+\mathrm{a})\left[1(\mathrm{a}\times \mathrm{a}-0)\right]=0 \\ \Rightarrow {\mathrm{a}}^2(3\mathrm{x}+\mathrm{a})=0 \\ \therefore \mathrm{x}=-\frac{\mathrm{a}}{3} \end{gathered}$$