Question

x³-x+3x²y+3xy²+y³-y

Answer 1

x³-x+3x²y+3xy²+y³-y

= (x+y)³  + (x+y)

= (x+y) [(x+y)²+1]

= (x+y) (x² + 2xy+y² +1)

Answer 2

x³ - x + 3x²y + 3xy² + y³ - y

= (x³ + 3x²y + 3xy² + y³) - y - x

= (x + y)³ - (x + y) = (x + y) [(x + y)² - 1]

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

Answer 3

x³ -x+3x²y+3xy²+y³ -y

=(x³ +3x²y+3xy²+y³) -(x+y)

=(x+y)³ -(x+y)=(x+y)(x +2xy+y -1)=(x+y)[(x+y)² -1]=(x+y)(x+y-1)(x+y+1)

k cho mik nha!