Question

!<phân tích thành nhân tử

a.(3x+1)^2-(x+1)^2

b.x^3+y^3+z^3-3xyz

Answer 1

a, \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\)

\(=2x.\left(4x+2\right)=4x\left(2x+1\right)\)

b, \(x^3+y^3+z^3-3xyz\)

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

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

\(=\left(x+y+z\right)\left[x^2+2xy+y^2-xz-yz+z^2-3xy\right]\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)