`((a+b+c)^(a+b)(a+b+c)^(b+c)(a+b+c)^(c+a))/([(a+b+c)^(a) (a+b+c)^(b) (a+b+c)^(c)]^(2))`

((a+b)^(2))/((b-c)(c-a))+((b+c)^(2))/((a-b)(c-a))+((c+a)^(2))/((a-b)(b-c))

Prove that |b c-a^2c a-b^2a b-c^2-b c+c a+a bb c-c a+a bb c+c a-a b(a+b)(a+c)(b+c)(b+a)(c+a)(c+b)|=3.(b-c)(c-a)(a-b)(a+b+c)(a b+b c+c a)

Prove: |((b+c)^2, a^2, b c) ,((c+a)^2, b^2 ,c a),( (a+b)^2, c^2, a b)|=(a-b)(b-c)(c-a)(a+b+c)(a^2+b^2+c^2) .

The expression (a-b)^3+\ (b-c)^3+\ (c-a)^3 can be factorized as (a) (a-b)(b-c)(c-a) (b) 3(a-b)(b-c)(c-a) (c) -3\ (a-b)(b-c)(c-a) (d) (a+b+c)(a^2+b^2+c^2-a b-b c-c a)

If a+b+c= 0 , then ((a+b)/(c ) + (b+c)/(a) + (c+ a)/(b)) ((c )/(a+b) + (b)/(a+c) + (a)/(b+c)) = ?