Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{x^3} + {y^3} + {z^3} - 3xyz\\
= {\left( {x + y} \right)^3} - 3xy\left( {x + y} \right) + {z^3} - 3xyz\\
= \left[ {{{\left( {x + y} \right)}^3} + {z^3}} \right] - \left( {3xy\left( {x + y} \right) + 3xyz} \right)\\
= {\left( {x + y + z} \right)^3} - 3\left( {x + y} \right)z\left( {x + y + z} \right) - 3xy\left( {x + y + z} \right)\\
= {\left( {x + y + z} \right)^3} - 3\left( {x + y + z} \right)\left( {xy + yz + zx} \right)\\
= \left( {x + y + z} \right)\left[ {{{\left( {x + y + z} \right)}^2} - 3\left( {xy + yz + zx} \right)} \right]\\
= \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right)\\
= \left( {x + y + z} \right).\frac{1}{2}.\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right]\\
\Rightarrow \frac{{{x^3} + {y^3} + {z^3} - 3xyz}}{{\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right]}} = \frac{1}{2}\left( {x + y + z} \right)
\end{array}\)
