Using properties of determinants, prove that  $\left|\begin{array}{ccc}{\mathrm{a}}^2+2\mathrm{a}& 2\mathrm{a}+1& 1\\ 2\mathrm{a}+1& \mathrm{a}+2& 1\\ 3& 3& 1\end{array}\right|={(\mathrm{a}-1)}^3.$

Let,

$$\begin{gathered} \mathrm{L}.\mathrm{H}.\mathrm{S}=\left|\begin{array}{ccc}{\mathrm{a}}^2+2\mathrm{a}& 2\mathrm{a}+1& 1\\ 2\mathrm{a}+1& \mathrm{a}+2& 1\\ 3& 3& 1\end{array}\right| \\ {\mathrm{R}}_2\to {\mathrm{R}}_2-{\mathrm{R}}_1, {\mathrm{R}}_3\to {\mathrm{R}}_3-{\mathrm{R}}_1 \\ =\left|\begin{array}{ccc}{\mathrm{a}}^2+2\mathrm{a}& 2\mathrm{a}+1& 1\\ 1-{\mathrm{a}}^2& -\mathrm{a}+1& 0\\ 3-{\mathrm{a}}^2-2\mathrm{a}& 3-2\mathrm{a}-1& 0\end{array}\right| \\ =\left|\begin{array}{ccc}{\mathrm{a}}^2+2\mathrm{a}& 2\mathrm{a}+1& 1\\ 1-{\mathrm{a}}^2& 1-\mathrm{a}& 0\\ 3-{\mathrm{a}}^2-2\mathrm{a}& 2-2\mathrm{a}& 0\end{array}\right| \end{gathered}$$

Expanding along ${\mathrm{C}}_3$

$$\begin{gathered} =1\left[\left(1-{\mathrm{a}}^2\right) \left(2-2\mathrm{a}\right)-\left(1-\mathrm{a}\right) \left(3-{\mathrm{a}}^2-2\mathrm{a}\right)\right] \\ =2\left(1-\mathrm{a}\right) \left(1-\mathrm{a}\right) \left(1+\mathrm{a}\right)-\left(1-\mathrm{a}\right) \left(3-{\mathrm{a}}^2-2\mathrm{a}\right) \\ =\left(1-\mathrm{a}\right) \left[2\left(1-{\mathrm{a}}^2\right)-3+{\mathrm{a}}^2+2\mathrm{a}\right] \\ =\left(1-\mathrm{a}\right) \left(2\mathrm{a}-{\mathrm{a}}^2-1\right) \\ ={\left(\mathrm{a}-1\right)}^3 \\ =\mathrm{RHS} \end{gathered}$$

Therefore, LHS= RHS have proven.