a) A = (a - b)3 + (b - c)3 + (c - a)3
Đặt : a - b = x ; b - c = y; c - a = z thì x + y + z = 0
Do đó: \(x^3+y^3+z^3=3xyz\)
Vậy A = \(3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
b) B = (a + b - 2c)3 + (b + c - 2a)3 + (c + a - 2b)3
Đặt : a + b - 2c = x ; b + c - 2a = y ; c + a - 2b = z
Thì x + y + z = 0 do đó \(x^3+y^3+z^3=3xyz\)
Vậy B = 3(a + b - 2c)(b + c - 2a)(c + a - 2b)
a) Ta có: \(A=\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3\)
\(=a^3-3a^2b+3ab^2-b^3+b^3-3b^2c+3bc^2-c^3+c^3-3c^2a+3ca^2-a^3\)
\(=-3\left(a^2b+ac^2-ab^2-bc^2+b^2c-a^2c\right)\)
\(=3\left[\left(a^2b-ab^2\right)+\left(ac^2-bc^2\right)-\left(a^2c-b^2c\right)\right]\)
\(=3\left[ab\left(a-b\right)+c^2\left(a-b\right)-c\left(a^2-b^2\right)\right]\)
\(=3\left[ab\left(a-b\right)+c^2\left(a-b\right)-c\left(a-b\right)\left(a+b\right)\right]\)
\(=3\left(a-b\right)\left[ab+c^2-c\left(a+b\right)\right]\)
\(=3\left(a-b\right)\left(ab+c^2-ca-cb\right)\)
\(=3\left(a-b\right)\left[\left(ab-ac\right)-\left(bc-c^2\right)\right]\)
\(=3\left(a-b\right)\left[a\left(b-c\right)-c\left(b-c\right)\right]\)
\(=3\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
b) 
