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

Lời giải

Ta có:

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

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

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

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

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

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