（x+y+z）대+（3x-2y-3z）대-（4x-y-2z）대分解因式

• (x+y+z)3+(3x-2y-3z)3-(4x-y-2z)3
  =(4x-y-2z)[(x+y+z)^2-2(x+y+z)(3x-2y-3z)+(3x-2y-3z)^2]-(4x-y-2z)3
  =(4x-y-2z)[(x+y+z)^2-2(x+y+z)(3x-2y-3z)+(3x-2y-3z)^2-(4x-y-2z)^2]
  =(4x-y-2z)[(x+y+z)((x+y+z)-(3x-2y-3z))+((3x-2y-3z)-(4x-y-2z))((3x-2y-3z)+(4x-y-2z))]
  =(4x-y-2z)[(x+y+z)(-2x+3y+4z)+(-x-y-z)(7x-3y-5z)]
  =(4x-y-2z)(x+y+z)[(-2x+3y+4z)-(7x-3y-5z)]
  =(4x-y-2z)(x+y+z)(-9x+6y+9z)
  =3(4x-y-2z)(x+y+z)(-3x+2y+3z)

(x+y+z)³+(3x-2y-3z)³-(4x-y-2Z)³
=(x+y+z+3x-2y-3z)[(x+y+z)²-(x+y+z)(3x-2y-3z)+(3x-2y-3z)²]-(4x-y-2z)³
=(4x-y-2z)[(x+y+z)²-(x+y+z)(3x-2y-3z)+(3x-2y-3z)²-(4x-y-2z)²]
=-3(4x-y-2z)(x+y+z)(3x-2y-3z)

把最后一个"(4x-y-2z)"拆成（x+y+z) +(3x-2y-3z)两项的和（x+y+z）=a (3x-2y-3z)=b 原始的化为a3+b3-(a+b)3 剩下的我就不算了 你懂的