Question

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

Answer 1

Considering

On applying C₁ → C₁ + C₂ + C₃

Take (a² + b² + c²), common

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

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

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

= R.H.S

Hence, proved.