Evaluate the following determinant:

(i) $\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+a& 1\\ 1& 1& 1+a\end{array}\right|=a^3+3a^2$

(ii) $\left|\begin{array}{ccc}{a}^2+2a& 2a+1& 1\\ 2a+1& a+2& 1\\ 3& 3& 1\end{array}\right|={\left(a-1\right)}^3$

(ii) To Prove: $\left|\begin{array}{ccc}{a}^2+2a& 2a+1& 1\\ 2a+1& a+2& 1\\ 3& 3& 1\end{array}\right|={\left(a-1\right)}^3$


$$\begin{gathered} \mathrm{LHS}=\left|\begin{array}{ccc}{a}^2+2a& 2a+1& 1\\ 2a+1& a+2& 1\\ 3& 3& 1\end{array}\right| \\ \mathrm{Applying}{R}_1\to R_1-R_2 \\ =\left|\begin{array}{ccc}{a}^2+2a-2a-1& 2a+1-a-2& 1-1\\ 2a+1& a+2& 1\\ 3& 3& 1\end{array}\right| \\ =\left|\begin{array}{ccc}{a}^2-1& a-1& 0\\ 2a+1& a+2& 1\\ 3& 3& 1\end{array}\right| \\ \mathrm{Taking}\left(a-1\right)\mathrm{common}\mathrm{from}{R}_1 \\ =\left(a-1\right)\left|\begin{array}{ccc}a+1& 1& 0\\ 2a+1& a+2& 1\\ 3& 3& 1\end{array}\right| \\ \mathrm{Applying}{R}_2\to R_2-R_3 \\ =\left(a-1\right)\left|\begin{array}{ccc}a+1& 1& 0\\ 2a+1-3& a+2-3& 1-1\\ 3& 3& 1\end{array}\right| \\ =\left(a-1\right)\left|\begin{array}{ccc}a+1& 1& 0\\ 2a-2& a-1& 0\\ 3& 3& 1\end{array}\right| \\ \mathrm{Taking}\left(a-1\right)\mathrm{common}\mathrm{from}{R}_2 \\ ={\left(a-1\right)}^2\left|\begin{array}{ccc}a+1& 1& 0\\ 2& 1& 0\\ 3& 3& 1\end{array}\right| \\ \mathrm{Expanding}\mathrm{through}{C}_3 \\ ={\left(a-1\right)}^2\left[1\left(1\left(a+1\right)-2\right)\right] \\ ={\left(a-1\right)}^2\left[1\left(a+1-2\right)\right] \\ ={\left(a-1\right)}^2\left(a-1\right) \\ ={\left(a-1\right)}^3=\mathrm{RHS} \end{gathered}$$


Hence, $\left|\begin{array}{ccc}{a}^2+2a& 2a+1& 1\\ 2a+1& a+2& 1\\ 3& 3& 1\end{array}\right|={\left(a-1\right)}^3$.