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

If a^(2) +b^(2) +c^(2) = ab + bc + ca then ((a+b)/(c ) + (b+c)/(a) + (c+a)/(b)) ((c )/(a+b) + (b)/(a+c) + (c )/(a+b)) =?

(a^(2)-b^(2)-2bc-c^(2))/(a^(2)+b^(2)+2ab-c^(2)) is equivalent to (a-b+c)/(a+b+c)( b) (a-b-c)/(a-b+c)(c)(a-b-c)/(a+b-c)(d)(a+b+c)/(a-b+c)

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) .

[[a,a^(2),(b+c)b,b^(2),(a+c)c,c^(2),(a+b)]]=(b-c)(c-a)(a-b)(a+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)

Given, a^(2) =b+c, b^(2)=c + a " & " c^(2) = a + b or (a^(2))/(b+c) = (b^(2))/(c+a) = (c^(2))/(a+b)=1 find (a)/(1+a) + (b)/(1+b) + (c )/(1+c) = ?

The determinant |[a^2, a^2-(b-c)^2,bc],[b^2,b^2-(c-a)^2,ca],[ c^2,c^2-(a-b)^2,ab]| is divisible by- a. a+b+c b. (a+b)(b+c)(c+a) c. a^2b^2c^2 d. (a-b)(b-c)(c-a)

Prove the following : |{:(a^(2),a,b+c),(b^(2),b,c+a),(c^(2),c,a+b):}|=-(a+b+c)(a-b)(b-c)(c-a)

|{:(a,b,c),(a^2,b^2,c^2),(b+c,c+a,a+b):}|=(a-b)(b-c)(c-a)(a+b+c)