Solution:

Option( B)
To find: Value of $\left|\begin{array}{ccc}x+y & x & x \\ 5 x+4 y & 4 x & 2 x \\ 10 x+8 y & 8 x & 3 x\end{array}\right|$
We have, $\left|\begin{array}{ccc}x+y & x & x \\ 5 x+4 y & 4 x & 2 x \\ 10 x+8 y & 8 x & 3 x\end{array}\right|$
Applying $R_{2} \rightarrow 2 R_{2}$
$\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}
x+y & x & x \\
10 x+8 y & 8 x & 4 x \\
10 x+8 y & 8 x & 3 x
\end{array}\right|$
Applying $R_{2} \rightarrow R_{2}-R_{3}$
$\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}
x+y & x & x \\
0 & 0 & x \\
10 x+8 y & 8 x & 3 x
\end{array}\right|$
Applying $\mathrm{R}_{1} \rightarrow 8 \mathrm{R}_{1}$
$\Rightarrow \frac{1}{2 \times 8}\left|\begin{array}{ccc}
8 x+8 y & 8 x & 8 x \\
0 & 0 & x \\
10 x+8 y & 8 x & 3 x
\end{array}\right|$
Applying $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}$
$\Rightarrow \frac{1}{16}\left|\begin{array}{ccc}
8 x+8 y & 8 x & 8 x \\
0 & 0 & x \\
2 x & 0 & -5 x
\end{array}\right|$
Expanding along $R_{2}$
$\begin{array}{l}
\Rightarrow \frac{1}{16}[x\{(2 x)(8 x)-(8 x+8 y)(0)\}] \\
\Rightarrow \frac{1}{16}\left[x\left\{16 x^{2}\right\}\right] \\
\Rightarrow x^{3}
\end{array}$