分解因式(x²+y²)³+(z²-x²)³-(y²

1个回答

  • (x²+y²)³+(z²-x²)³-(y²+z²)³

    = [(x²+y²)+(z²-x²)][(x²+y²)²+(z²-x²)²-(x²+y²)(z²-x²)]-(y²+z²)³

    = (y²+z²)[(x²+y²)²+(z²-x²)²-(x²+y²)(z²-x²)]-(y²+z²)³

    = (y²+z²)[(x²+y²)²+(z²-x²)²-(x²+y²)(z²-x²)-(y²+z²)²]

    = (y²+z²)[(x²+y²-y²-z²)(x²+y²+y²+z²)+(z²-x²)(z²-x²-x²-y²)]

    =(y²+z²)[(x²-z²)(x²+2y²+z²)+(z²-x²)(z²-2x²-y²)]

    =(y²+z²)[(x²-z²)(x²+2y²+z²-z²+2x²+y²]

    =3(y²+z²)(x²-z²)(x²+y²)