Question

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

Answer 1

$$\begin{gathered} ∆=\left|\begin{array}{ccc}1+a& 1& 1\\ 1& 1+a& 1\\ 1& 1& 1+a\end{array}\right| \\ =1+a\left|\begin{array}{cc}1+a& 1\\ 1& 1+a\end{array}\right|-1\left|\begin{array}{cc}1& 1\\ 1& 1+a\end{array}\right|+1\left|\begin{array}{cc}1& 1+a\\ 1& 1\end{array}\right|\left[\mathrm{Expanding}\right] \\ =(1+a)\left[(1+a{)}^2-1\right]-1(1+a-1)+(1-1-a) \\ =(1+a)[1+a^2+2a-1]-a-a \\ =1+a+a^2+a^3+2a+2a^2-2a \\ =a^3+3a^2 \end{gathered}$$