`(a^3+b^3-c^3+3abc)/(a+b-c)=`

If a = 25,b=15, c=-10, then the value of (a^3+b^3+c^3-3abc)/((a-b)^2+(b-c)^2+(c-a)^2) is यदि a = 25,b=15, c=-10 तो (a^3+b^3+c^3-3abc)/((a-b)^2+(b-c)^2+(c-a)^2) का मान है-

If a=-1,b=2,c=3," then"(a^(3)+b^(3)+c^(3)-3abc)/((a-b)^(2)+(b-c)^(2)+(c-a)^(2))=

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

if abc!=0 and if |[a,b,c],[b,c,a],[c,a,b]|=0 then (a^3+b^3+c^3)/(abc)=

If abc!=0 and if |(a,b,c),(b,c,a),(c,a,b)|=0 then (a^(3)+b^(3)+c^(3))/(abc)=

In the formula, a^3+b^3+c^3-3 abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) if a + b + c = 0 , then show that a^3 + b^3 + c ^3 = 3 abc

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

Let's write the value of (a + b + c) by calculation if a^3 + b^3 + c^3 = 3abc (a ne b ne c).

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