Using properties of determinants prove that: |((b+c)^2 a^2 bc) ((c+a)^2 b^2,ca) ((a+b)^2, c^2, ab)| = (a^2 + b^2 +c^2) (a-b) (b-c) (c-a) (a+b+c)

Question

Using properties of determinants prove that:

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

|((b+c)² a² bc) ((c+a)² b²,ca) ((a+b)², c², ab)| = (a² + b² +c²) (a-b) (b-c) (c-a) (a+b+c)

Answer 1

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

= (a² + b² + c²)(a - b)(b - c){0 + 0 - a(a + b) + c(b + c)} [expansion by first row] 

= (a² + b² + c²)(a - b)(b - c){0 + 0 - a(a + b) + c(b + c)} 

= (a² + b² + c²)(a - b)(b - c)( - a² - ba + bc + c²) 

= (a² + b² + c²)(a - b)(b - c){(c² - a²) + b(c - a)} 

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