Question

Phân tích đa thức thành nhân tử 
x³+y³+3y²+3y+1

Answer 1

\(x^3+y^3+3y^2+3y+1\\ =x^3+\left(y+1\right)^3\\ =\left(x+y+1\right)\left[x^2-x\left(y+1\right)+\left(y+1\right)^2\right]\\ =\left(x+y+1\right)\left(x^2-xy-x+y^2+2y+1\right)\\ =\left(x+y+1\right)\left(x^2+y^2+2y+1-xy-x\right)\)

Answer 2

=(x^3+3y+3y^2+y^3)+1

=(x+y)^3+1

=(x+y+1).[(x+y)^2+(x+y)+1]