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In Radius

Prove that:1 ...

Question

Prove that :

$|\begin{matrix} 1 & a^{2} + b c & a^{3} \\ 1 & b^{2} + c a & b^{3} \\ 1 & c^{2} + a b & c^{3} \end{matrix}| = - (a - b) (b - c) (c - a) (a^{2} + b^{2} + c^{2})$

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Solution

$Let LHS = Δ = |1 a^{2} + bc a^{3} 1 b^{2} + ca b^{3} 1 c^{2} + ab c^{3}| \Rightarrow Δ = |0 (a^{2} + bc) - (b^{2} + ca) a^{3} - b^{3} 0 (b^{2} + ca) - (c^{2} + ab) b^{3} - c^{3} 1 c^{2} + ab c| [Applying R_{1} \to R_{1} - R_{2} and R_{2} \to R_{2} - R_{3}]$

$= |0 a^{2} - b^{2} - ca + bc a^{3} - b^{3} 0 b^{2} - c^{2} - ab + ca b^{3} - c^{3} 1 c^{2} + ab c^{3}| = |0 (a - b) (a + b - c) (a - b) (a^{2} + ab + b^{2}) 0 (b - c) (b + c - a) (b - c) (b^{2} + bc + a^{2}) 1 c^{2} + ab c^{3}|$

$= (a - b) (b - c) |0 a + b - c a^{2} + ab + b^{2} 0 (b + c - a) (b^{2} + bc + c^{2}) 1 c^{2} + ab c^{3}| [Taking out (a - b) common from R_{1} and (b - c) from R_{2}] = (a - b) (b - c) |0 a + b - c a^{2} + ab + b^{2} 0 (b + c - a) - (a + b - c) (b^{2} + bc + c^{2}) - (a^{2} + ab + b^{2}) 1 c^{2} + ab c^{3}| [Applying R_{2} \to R_{2} - R_{1}]$

$= (a - b) (b - c) |0 a + b - c a^{2} + ab + b^{2} 0 2 (c - a) b (c - a) + (c^{2} - a^{2}) 1 c^{2} + ab c^{3}| = (a - b) (b - c) (c - a) |0 a + b - c a^{2} + ab + b^{2} 0 2 a + b + c 1 c^{2} + ab c^{3}| = (a - b) (b - c) (c - a) \times \{1 \times |a + b - c a^{2} + ab + b^{2} 2 a + b + c|\} [Expanding along C_{1}]$

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

Hence proved.

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|\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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