Using `a^3+b^3+c^3-3a b c=(a+b+c)(a^2+b^2+c^2-a b-b c-c a)`

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)

(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

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

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

In the formula, a^3+b^3+c^3-3 abc = 1/2 (a+b+c){(a-b)^2+(b-c)^2+(c-a)^2} , if a + b + c != 0 , Show that if a^3 +b^3+c^3= 3abc impliesa=b=c

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

a^(3)+b^(3)+c^(3)-3abc=k(a+b+c)[(a-b)^(2)+(b-c)^(2)+(c-a)^(2)] , then k=

If a= 25, b= 15, c= - 10 then (a^(3) + b^(3) + c^(3) - 3abc)/((a-b)^(2) + (b-c)^(2) + (c-a)^(2))

If A=|(1,1,1),(a,b,c),(a^3,b^3,c^3)|, B=|(1,1,1),(a^2,b^2,c^2),(a^3,b^3,c^3)|, C=|(a,b,c),(a^2,b^2,c^2),(a^3,b^3,c^3)| , then which relation is correct :