सिद्ध कीजिए:
`(a+b)^(3)+(b+c)^(3)+(c+a)^(3)-3(a+b)(b+c)(c+a)=2(a^(3)+b^(3)+c^(3)-3abc)`

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)-3abc)

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)

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)

If 2s=a+b+c, prove that (s-a)^(3)+(s-b)^(3)+(s-c)^(3)-3(s-a)(s-b)(s-c)=(1)/(2)(a^(3)+b^(3)+c^(3)-3abc)

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

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

If a + b + c = 0 , show that a^(3) + b^(3) + c^(3) = 3abc The following are the steps involved in showing the above result. Arrange them in sequential order (A) a^(3) + b^(3) + 3ab (-c) = -c^(3) (B) (a + b)^(3) = (-c)^(3) (C) a + b + c = 0 rArr a + b = -c (D) a^(3) + b^(3) + 3ab (a +b) = -c^(3) (E) a^(3) + b^(3) + c^(2) = 3abc

Prove (a+2b)^3+(b+2c)^3+(c+2a)^3+3(a+3b+2c)(b+3c+2a)(c+3a+2b)=27(a+b+c)^3

If 2s = a + b + c , Then prove that (s-a)^3+(s-b)^3+(s-c)^3-3(s-a)(s-b)(s-c) = 1/2(a^3+b^3+c^3-3abc)