x^3 - x + 3x^2y + 3xy^2 + y^3 - y

=x³+y³+3x²y+3xy²-x-y

=(x+y)(x²-xy+y²)+3xy(x+y)-(x+y)

=(x+y)(x²-xy+y²+3xy-1)

=(x+y)(x²+2xy+y²-1)

=(x+y)[(x+y)²-1]

=(x+y)(x+y-1)(x+y+1)

x^2 + 5x - 6

=x²-x+6x-6

=x.(x-1)+6.(x-1)

=(x-1)(x+6)

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)

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)

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

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

\(=6x^2+5x-3xy\)

\(=x\left(6x+5-3y\right)\)

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

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

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

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

HỌC TỐT NHA!

ta có:
x³ + y³ + z³ - 3xyz
= (x+y)³ - 3xy(x-y) + z³ - 3xyz
= [(x+y)³ + z³] - 3xy(x+y+z)
= (x+y+z)³ - 3z(x+y)(x+y+z) - 3xy(x-y-z)
= (x+y+z)[(x+y+z)² - 3z(x+y) - 3xy]
= (x+y+z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= (x+y+z)(x² + y² + z² - xy - xz - yz)

- Hình như đề của u sai hay sao á :)))