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Invertible Element Binary Operation

Prove the fol...

Question

Prove the following identities:

$|\begin{matrix} a^{3} & 2 & a \\ b^{3} & 2 & b \\ c^{3} & 2 & c \end{matrix}| = 2 (a - b) (b - c) (c - a) (a + b + c) \begin{matrix} \end{matrix}$

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Solution

$LHS = |\begin{matrix} a^{3} & 2 & a \\ b^{3} & 2 & b \\ c^{3} & 2 & c \end{matrix}| = |\begin{matrix} a^{3} & 2 & a \\ b^{3} - a^{3} & 0 & b - a \\ c^{3} - a^{3} & 0 & c - a \end{matrix}| [Applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}] = - (a - b) (c - a) |\begin{matrix} a^{3} & 2 & a \\ b^{2} + a^{2} + a b & 0 & 1 \\ c^{2} + a^{2} + a c & 0 & 1 \end{matrix}| [Taking (b - a) common from R_{2} and (c - a) common from R_{3}] = - (a - b) (c - a) |\begin{matrix} a^{3} & 2 & a \\ b^{2} - c^{2} + a b - a c & 0 & 0 \\ c^{2} + a^{2} + a c & 0 & 1 \end{matrix}| [Applying R_{2} \to R_{2} - R_{3}] = - (a - b) (c - a) |\begin{matrix} a^{3} & 2 & a \\ (b - c) (a + b + c) & 0 & 0 \\ c^{2} + a^{2} + a c & 0 & 1 \end{matrix}| = - (a - b) (c - a) (b - c) (a + b + c) |\begin{matrix} a^{3} & 2 & a \\ 1 & 0 & 0 \\ c^{2} + a^{2} + a c & 0 & 1 \end{matrix}| [Taking (b - c) (a + b + c) common from R_{2}] = - (a - b) (c - a) (b - c) (a + b + c) (- 2) [Expanding along second column] = 2 (a - b) (c - a) (b - c) (a + b + c) = RHS ∴ |\begin{matrix} a^{3} & 2 & a \\ b^{3} & 2 & b \\ c^{3} & 2 & c \end{matrix}| = 2 (a - b) (b - c) (c - a) (a + b + c) \begin{matrix} \end{matrix}$

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