Question

Phân tích đa thức sau thành nhân tử

\(x^3-3x^2+3x-1-y^3\)

Answer 1

\(x^3-3x^2+3x-1-y^3\)

\(=\left(x-1\right)^3-y^3\)

\(=\left(x-1-y\right)\left[\left(x-1\right)^2+y\left(x-1\right)+y^2\right]\)

\(=\left(x-y-1\right)\left[\left(x-1\right)\left(x-1+y\right)+y^2\right]\)

Answer 2

\(x^3-3x^2+3x-1-y^3\\ =\left(x-1\right)^3-y^3\\ =\left(x-1-y\right)\text{[ (x-1)^2+y(x-1)+y^2}\)

\(=\left(x-y-1\right)\left[\left(x-1\right)\left(x-1+y\right)+y^2\right]\)