Rút gọn phân thức sau: \[\frac{{{x^3} + {y^3} + {z^3} - 3xyz}}{{{{(x - y)}^2} + {{(y - z)}^2} + {{(x - z)}^2}}}\].

Lời giải:

Ta có: \[\frac{{{x^3} + {y^3} + {z^3} - 3xyz}}{{{{(x - y)}^2} + {{(y - z)}^2} + {{(x - z)}^2}}}\]

= \[\frac{{{{(x + y)}^3} - 3{x^2}y - 3x{y^2} - 3xyz + {z^3}}}{{{x^2} - 2xy + {y^2} - 2yz + {z^2} + {x^2} - 2xz + {z^2}}}\]

= \[\frac{{{\rm{[}}{{(x + y)}^3} + {z^3}{\rm{]}} - 3xy(x + y + z)}}{{2{x^2} + 2{y^2} + 2{z^2} - 2xy - 2yz - 2xz}}\]

= \(\frac{{(x + y + z){\rm{[}}{{(x + y)}^2} - z(x + y) + {z^2}{\rm{]}} - 3xy(x + y + z)}}{{2({x^2} + {y^2} + {z^2} - xy - yz - xz)}}\)

= \(\frac{{(x + y + z)({x^2} + 2xy - zx - zy + {z^2} - 3xy)}}{{2({x^2} + {y^2} + {z^2} - xy - yz - xz)}}\)

= \(\frac{{(x + y + z)({x^2} + {y^2} + {z^2} - xy - yz - xz)}}{{2({x^2} + {y^2} + {z^2} - xy - yz - xz)}}\)

= \[\frac{{x + y + z}}{2}\]= \[\frac{1}{2}(x + y + z)\].