Question

[phan thị minh anh]

phân tích đa thức thành nhân tử : 

a. \(\left(a+b+c\right)^3-a^3-b^3-c^3\)

b.\(8\left(x+y+z\right)^3-\left(x+y\right)^3-\left(y+z\right)^3-\left(x+z\right)^3\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(\left(a+b+c\right)^3-a^3-b^3-c^3\)

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

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

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

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

\(=3\left(b+c\right)\left(a+b\right)\left(a+c\right)\)

b: \(=\left(2x+2y+2z\right)^3-\left(x+y\right)^3-\left[\left(y+z\right)^3+\left(x+z\right)^3\right]\)

\(=\left(x+y+2z\right)\left[\left(2x+2y+2z\right)^2+2\left(x+y+z\right)\left(x+y\right)+\left(x+y\right)^2\right]-\left(x+y+2z\right)\left[\left(y+z\right)^2-\left(y+z\right)\left(x+z\right)+\left(x+z\right)^2\right]\)

\(=3\left(x+y+2z\right)\left(x+z+2y\right)\left(y+z+2x\right)\)