निम्न व्यंजको के गुणनखण्ड कीजिये:
`(a-b+c)^(2)+(b-c+a)^(2)+2(a-b+c)(b-c+a)`

|(a-b-c,2b,2c),(2a,b-c-a,2c),(2a,2b,c-a-b)|=

Prove that =|1 1 1a b c b c+a^2a c+b^2a b+c^2|=2(a-b)(b-c)(c-a)

Prove that =|1 1 1a b c b c+a^2a c+b^2a b+c^2|=2(a-b)(b-c)(c-a)

Value of |(1,a,a^2),(1,b,b^2),(1,c,c^2)| is (A) (a-b)(b-c)(c-a) (B) (a^2-b^2)(b^2-c^2)(c^2-a^2) (C) (a-b+c)(b-c+a)(c+a-b) (D) none of these

Value of |(1,a,a^2),(1,b,b^2),(1,c,c^2)| is (A) (a-b)(b-c)(c-a) (B) (a^2-b^2)(b^2-c^2)(c^2-a^2) (C) (a-b+c)(b-c+a)(c+a-b) (D) none of these

(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)

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)