if `(a+b+c)^3=a^3+b^3+c^3` then `(a+b)(b+c)(c+a)` equal to ``
A.3` `
B.1` `
C.0` `
D.-1``

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

If a+b+c=0, then a^(3)+b^(3)+c^(3) is equal to

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

Factorise : (a+b)^3+(b+c)^3+(c+a)^3-3(a+b)(b+c)(c+a)

Factorise : (a+b)^3+(b+c)^3+(c+a)^3-3(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)

If a/b+b/a=1, then a^3+b^3=? (a) 1 (b) -1 (c) 1/2 (d) 0

If a b c = 0, then ({(x^a)^b}^c)/({(x^b)^c}^a) = (a)3 (b) 0 (c) -1 (d) 1

If a/b+b/a=\ -1, then a^3-b^3=? (a)1 (b) -1 (c) 1/2 (d) 0

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}