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Question
\[\begin{vmatrix}1 + a & 1 & 1 \\ 1 & 1 + a & a \\ 1 & 1 & 1 + a\end{vmatrix} = a^3 + 3 a^2\]
Solution
\[∆ = \begin{vmatrix}1 + a & 1 & 1 \\ 1 & 1 + a & 1 \\ 1 & 1 & 1 + a\end{vmatrix}\]
\[ = 1 + a \begin{vmatrix}1 + a & 1 \\ 1 & 1 + a\end{vmatrix} - 1\begin{vmatrix}1 & 1 \\ 1 & 1 + a\end{vmatrix} + 1\begin{vmatrix}1 & 1 + a \\ 1 & 1\end{vmatrix} \left[\text{ Expanding }\right]\]
\[ = (1 + a)\left[ (1 + a )^2 - 1 \right] - 1(1 + a - 1) + (1 - 1 - a)\]
\[ = (1 + a)[1 + a^2 + 2a - 1] - a - a\]
\[ = 1 + a + a^2 + a^3 + 2a + 2 a^2 - 2a\]
\[ = a^3 + 3 a^2\]
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