Question

\(\begin{vmatrix} a^2 & a^2 - (b - c)^2 & bc \\[0.3em] b^2 &b^2 - (c - a)^2 & ca \\[0.3em] c^2 &c^2 - (a - b)^2 &ab \end{vmatrix}\) = (a - b)(b - c)(c - a)(a + b + c)(a² + b² + c²)

Answer 1

Considering

On applying C₂ → C₂ - 2C₁ - 2C₃

Take -(a² + b² + c²) common from C₂

On applying R₂ → R₂ - R_1, and R₃ → R₃ - R₁

= – (a² + b² + c²) (a – b) (c – a) [(– (b + a)) (– b) – (c) (c + a)]

= (a – b) (b – c) (c – a) (a + b + c) (a² + b² + c²)

= R.H.S

Hence, proved.