Question

phan tich da thuc thanh nhan tu

X^3_X+3X²Y+3XY²+Y^3-Y

Answer 1

x³-x+3x²y+3xy²+y³-y

=x²(x-1)+3(x²y+xy²)+y²(y-1)

=x²(x-1)+3(x².y+y².x)+y²(y-1)

=x²(x-1)+3{[x(x+1)+y(y+1)]}+y²(y-1)

=x²(x-1)+3.x(x+1)+3.y(y+1)+y²(y-1)

=x²(x-1)+2x²+3.x(x+1)+3.y(y+1)+y²(y-1)+2y²-2x²-2y²

=x²(x+1)+3.x(x+1)+3.y(y+1)+y²(y+1)-2x²-2y²

=(x²+3)(x+1)+(y²+3)(y+1)-2(x²+y²)

Answer 2

ta có : (x*3+3x*2y+3xy*2+y*3)-(x+y)

=(x+y)*3-(x+y)

=(x+y)((X+Y)*2-1)

(x+y)(x+y+1)(x+Y-1)

Answer 3

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

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