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

We have to prove,
$= \left| {\begin{array}{cccccccccccccccccccc}{{a^2} + 2a}&{2a + 1}&1\\{2a + 1}&{a + 2}&1\\3&3&1\end{array}} \right| = {(a - 1)^3}$

$\therefore$ ${\rm{LHS}} = \left| {\begin{array}{cccccccccccccccccccc}{{a^2} + 2a}&{2a + 1}&1\\{2a + 1}&{a + 2}&1\\3&3&1\end{array}} \right|$

$= \left| {\begin{array}{cccccccccccccccccccc}{{a^2} + 2a - 2a - 1}&{2a + 1 - a - 2}&0\\{2a + 1 - 3}&{a + 2 - 3}&0\\3&3&1\end{array}} \right|$

and $\left. {{R_2} \to {R_2} - {R_3}} \right]$
$= \left| {\begin{array}{cccccccccccccccccccc}{(a - 1)(a + 1)}&{(a - 1)}&0\\{2(a - 1)}&{(a - 1)}&0\\3&3&1\end{array}} \right|$

$= {(a - 1)^2}\left| {\begin{array}{cccccccccccccccccccc}{(a + 1)}&1&0\\2&1&0\\3&3&1\end{array}} \right|$

[taking $(a - 1)$ common from ${R_1}$

and ${R_2}$ each] $= {(a - 1)^2}[1(a + 1) - 2] = {(a - 1)^3}$
$= {\rm{RHS}}$