Using properties of determinants prove that $\begin{vmatrix} a & a+b & a+b+c \\ 2a & 3a+2b & 4a+3b+2c \\ 3a & 6a+3b & 10a+6b+3c \end{vmatrix} = a^3$
Let $\Delta =$
$
\begin{vmatrix}
a & a+b & a+b+c \\
2a & 3a+2b & 4a+3b+2c \\
3a & 6a+3b & 10a+6b+3c
\end{vmatrix}
$
$R_2 \rightarrow R_2 - 2R_1, R_3 \rightarrow R_3 - 3R_1 \implies \Delta =$
$
\begin{vmatrix}
a & a+b & a+b+c \\
0 & a & 2a+b \\
0 & 3a & 7a+3b
\end{vmatrix}
$
$R_3 \rightarrow R_3 - 3R_2 \implies \Delta =$
$
\begin{vmatrix}
a & a+b & a+b+c \\
0 & a & 2a+b \\
0 & 0 & a
\end{vmatrix}
$
Expanding along $R_1$ gives
$\Delta = a \times a \times a = a^3$