The value of `[{(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3}/{(a-b)^3+(b-c)^3+(c-a)^3}] =`
(1) `3(a+b)(b+c)(c+a)`
(2) `3(a-b)(b-c)(c-a)`
(3) `(a+b)(b+c)(c+a)`
(4) `1`

Simplify: ((a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3)/((a-b)^3+(b-c)^3+(c-a)^3)

Simplify: ((a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3)/((a-b)^3+(b-c)^3+(c-a)^3)

Simplify: ((a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3)/((a-b)^3+(b-c)^3+(c-a)^3)

Simplify: ((a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3)/((a-b)^3+(b-c)^3+(c-a)^3)

(a-b)^3 + (b-c)^3 + (c-a)^3=? (a) (a+b+c)(a^2+b^2+c^2-ab-bc-ac) (b) 3(a-b)(b-c)(c-a) (c) (a-b)(b-c)(c-a) (d)none of these

Simplify ((a^(2)-b^(2))^(3)+(b^(2)-c^(2))^(3)+(c^(2)-a^(2))^(3))/((a-b)^(3)+(b-c)^(3)+(c-a)^(3))

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)

The value of the determinant |(1,a,a^2-bc),(1,b,b^2-ca),(1,c,c^2-ab)| is (A) (a+b+c),(a^2+b^2+c^2) (B) a^3+b^3+c^3-3abc (C) (a-b)(b-c)(c-a) (D) 0

Prove that a^3+b^3+c^3-3abc=1/2(a+b+c){(a-b)^2+(b-c)^2+(c-a)^2}

Prove that : (a+b)^3+(b+c)^3+(c+a)^3-3(a+b)(b+c)(c+a)=2(a^3+b^3+c^3-3a b c)