Bài 1 :
$x^3-3x^2y+3xy^2-y^3-z^3$
$=(x-y)^3 - z^3 $
$= (x - y -z)[ (x-y)^2 +(x-y)z + z^2 ]$
Bài 2 :
$ (2x-3)^6 + 2.(2x-3)^2 = 0 $
$ (2x-3)^2 [(2x-3)^4 + 2 ] = 0$
⇒$(2x-3)^2 = 0$
$2x - 3 = 0$
$2x = 3 $
$x = \dfrac{3}{2}$ => lấy
⇒$ (2x-3)^4 + 2 = 0$
$(2x-3)^4 =-2 $$=> bỏ $