Question:
$\left|\begin{array}{ccc}a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1\end{array}\right|=(a-1)^{3}$
Solution:
Given, $\left|\begin{array}{ccc}a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1\end{array}\right|$
[Applying $R_{1} \rightarrow R_{1}-R_{2}$ and $R_{2} \rightarrow R_{2}-R_{3}$ ]
$=\left|\begin{array}{ccc}a^{2}-1 & a-1 & 0 \\ 2 a-2 & a-1 & 0 \\ 3 & 3 & 1\end{array}\right|$
Now,
[Taking $(a-1)$ common from $R_{1}$ and $R_{2}$ ]
$(a-1)^{2}\left|\begin{array}{ccc}a+1 & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1\end{array}\right|$
Finally,
[Expanding along $R_{3}$ ]
$=(a-1)^{2}[1 \cdot(a+1)-2]=(a-1)^{3}$