Question

Phân tích sau thành nhân tử: \(a^3+b^3+c^3-3abc\)

b. \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)

Answer 1

a. \(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc\)

\(=[\left(a+b\right)^3+c^3]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)[\left(a+b\right)^2-c\left(a+b\right)+c^2]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

b. Đặt x - y = a, y - z = b, z - x = c thì a + b + c = 0

Do đó theo câu a ta có: \(a^3+b^3+c^3-3abc=0\Rightarrow a^3+b^3+c^3=3abc\)

\(\Rightarrow\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)