Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
{\left( {b - c} \right)^3} + {\left( {c - a} \right)^3} + {\left( {a - b} \right)^3}\\
 = \left[ {{{\left( {b - c} \right)}^3} + {{\left( {c - a} \right)}^3}} \right] + {\left( {a - b} \right)^3}\\
 = \left[ {\left( {b - c} \right) + \left( {c - a} \right)} \right].\left[ {{{\left( {b - c} \right)}^2} - \left( {b - c} \right).\left( {c - a} \right) + {{\left( {c - a} \right)}^2}} \right] + {\left( {a - b} \right)^3}\\
 = \left( {b - a} \right).\left[ {\left( {{b^2} - 2bc + {c^2}} \right) - \left( {bc - ba - {c^2} + ca} \right) + \left( {{c^2} - 2ca + {a^2}} \right)} \right] + {\left( {a - b} \right)^3}\\
 = \left( {b - a} \right).\left( {{b^2} - 2bc + {c^2} - bc + ba + {c^2} - ca + {c^2} - 2ca + {a^2}} \right) + {\left( {a - b} \right)^3}\\
 = \left( {b - a} \right)\left( {{a^2} + {b^2} + 3{c^2} - 3bc - 3ca + ab} \right) - {\left( {b - a} \right)^3}\\
 = \left( {b - a} \right).\left[ {\left( {{a^2} + {b^2} + 3{c^2} - 3bc - 3ca + ab} \right) - {{\left( {b - a} \right)}^2}} \right]\\
 = \left( {b - a} \right)\left( {{a^2} + {b^2} + 3{c^2} - 3bc - 3ca + ab - {b^2} + 2ba + {a^2}} \right)\\
 = \left( {b - a} \right).\left( {3{c^2} - 3bc - 3ca + 3ab} \right)\\
 = 3.\left( {b - a} \right).\left( {{c^2} - bc - ca + ab} \right)\\
 = 3.\left( {b - a} \right).\left[ {\left( {{c^2} - bc} \right) + \left( {ab - ca} \right)} \right]\\
 = 3.\left( {b - a} \right).\left[ {c\left( {c - b} \right) + a.\left( {b - c} \right)} \right]\\
 = 3.\left( {b - a} \right).\left( {c - b} \right)\left( {c - a} \right)\\
 = 3.\left( {a - b} \right).\left( {b - c} \right).\left( {c - a} \right)\\
b,\\
{x^3} + {y^3} + {z^3} - 3xyz\\
 = \left( {{x^3} + 3{x^2}y + 3x{y^2} + {y^3}} \right) + {z^3} - \left( {3{x^2}y + 3x{y^2} + 3xyz} \right)\\
 = {\left( {x + y} \right)^3} + {z^3} - 3xy\left( {x + y + z} \right)\\
 = \left[ {\left( {x + y} \right) + z} \right].\left[ {{{\left( {x + y} \right)}^2} - \left( {x + y} \right).z + {z^2}} \right] - 3xy.\left( {x + y + z} \right)\\
 = \left( {x + y + z} \right).\left( {{x^2} + 2xy + {y^2} - xz - yz + {z^2}} \right) - 3xy\left( {x + y + z} \right)\\
 = \left( {x + y + z} \right)\left[ {\left( {{x^2} + 2xy + {y^2} - xz - yz + {z^2}} \right) - 3xy} \right]\\
 = \left( {x + y + z} \right).\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right)
\end{array}\)