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Absolute Value Function

Prove that:a ...

Question

Prove that :

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

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Solution

$Let LHS = Δ = |a^{2} a^{2} - {(b - c)}^{2} bc b^{2} b^{2} - {(c - a)}^{2} ca c^{2} c^{2} - {(a - b)}^{2} ab|$

$\Rightarrow ∆ = |a^{2} - {(b - c)}^{2} bc b^{2} - {(c - a)}^{2} ca c^{2} - {(a - b)}^{2} ab| [Applying C_{2} \to C_{2} - C_{1}] = (- 1) |a^{2} {(b - c)}^{2} bc b^{2} {(c - a)}^{2} ca c^{2} {(a - b)}^{2} ab| = - |a^{2} b^{2} + c^{2} bc b^{2} c^{2} + a^{2} ca c^{2} a^{2} + b^{2} ab| [Applying C_{2} \to C_{2} - 2 C_{1}]$

$= - |a^{2} + b^{2} + c^{2} b^{2} + c^{2} bc b^{2} + c^{2} + a^{2} c^{2} + a^{2} ca c^{2} + a^{2} + b^{2} a^{2} + b^{2} ab| [Applying C_{1} \to C_{1} + C_{2}] = - (a^{2} + b^{2} + c^{2}) |1 b^{2} + c^{2} bc 1 c^{2} + a^{2} ca 1 a^{2} + b^{2} ab|$

$= - (a^{2} + b^{2} + c^{2}) |1 b^{2} + c^{2} bc 0 (c^{2} + a^{2}) - (b^{2} + c^{2}) ca - bc 0 (a^{2} + b^{2}) - (b^{2} + c^{2}) ab - bc| [Applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}] = ((a^{2} + b^{2} + c^{2})) |1 b^{2} + c^{2} bc 0 a^{2} - b^{2} c (a - b) 0 a^{2} - c^{2} b (a - c)| = - (a^{2} + b^{2} + c^{2}) (a - b^{}) (a - c) |1 b^{2} + c^{2} bc 0 a + b^{} c 0 a^{} + c^{} b| [Taking (a - b) common from R_{2} and (a - c) common from R_{3}] = (a^{2} + b^{2} + c^{2}) (a - b^{}) (c - a) \times \{1 \times |a + b^{} c a^{} + c^{} b|\} [∵ (c - a) = - (a - c)] [Expanding along C_{1}] = (a^{2} + b^{2} + c^{2}) (a - b^{}) (c - a) (ab + b^{2} - ac - c^{2})$

$= (a^{2} + b^{2} + c^{2}) (a - b) (c - a) \{a (b - c) + (b + c) (b - c)\} = (a - b) (c - a) (b - c) (a + b + c) (a^{2} + b^{2} + c^{2})$

= RHS

Hence proved.

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