Show that: `|a a+b a+b+c2a3a+2b4a+3b+2c3a6a+3b 10 a+6b+3c|=a^3` .

Answer

Proved

Explanation

$$=\left|\begin{array}{ccc}a & a+b & a+b+c \\ 2 a & 3 a+2 b & 4 a+3 b+2 c \\ 3 a & 6 a+3 & 10 a+6 b+3 c\end{array}\right|$$
$$=\left|\begin{array}{ccc}a & a & a+b+c \\ 2 a & 3 a & 4 a+3 b+2 c \\ 3 a & 6 a & 10 a+6 b+3 c\end{array}\right|+\left|\begin{array}{ccc}a & b & a+b+c \\ 2 a & 2 b & 4 a+3 b+2 c \\ 3 a & 3 b & 10 a+6 b+3 c\end{array}\right|$$
`[c_{1}=c_{2}` in second det.]
$$ =a^{2} =\left | \begin{array}{ccc}1 & 1 & a+b+c \\ 2 & 3 & 4 a+3 b+2 c \\ 3 & 6 & 10 a+6 b+3 c\end{array} \right |$$
$$=a^{2}\left|\begin{array}{ccc}1 & 1 & a \\ 2 & 3 & 4 a \\ 3 & 6 & 10 a\end{array}\right|+a^{2}\left|\begin{array}{lll}1 & 1 & b \\ 2 & 3 & 3 c \\ 3 & 6 & 60\end{array}\right|+a^{2}+k^{c}\left|\begin{array}{lll}1 & 1 & c \\ 2 & 3 & 2 c \\ 3 & 6 & 3 c\end{array}\right|$$
$$=a^{3} a\left|\begin{array}{lll}1 & 1 & 1 \\ 2 & 3 & 4 \\ 3 & 6 & 10\end{array}\right|+a^{3} \cdot|| \begin{array}{lll}1 & 1 & 1 \\ 2 & 3 & 3 \\ 3 & 6 & 6\end{array}\left|+a^{2} \cdot c\right| \begin{array}{lll}1 & 1 & 1 \\ 2 & 3 & 2 \\ 3 & 6 & 3\end{array} \mid$$
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