Đáp án:

$\begin{array}{l}
a){\left( {a - b} \right)^3} + {\left( {b - c} \right)^3} + {\left( {c - a} \right)^3}\\
 = \left( {a - b + b - c} \right)\left( {{{\left( {a - b} \right)}^2} - \left( {a - b} \right)\left( {b - c} \right) + {{\left( {b - c} \right)}^2}} \right) - {\left( {a - c} \right)^3}\\
 = \left( {a - c} \right).\left( {{a^2} - 2ab + {b^2} - ab + ac + {b^2} - bc + {b^2} - 2bc + {c^2}} \right)\\
 - {\left( {a - c} \right)^3}\\
 = \left( {a - c} \right).\left( {{a^2} + 3{b^2} + {c^2} - 3ab + ac - 3bc - {{\left( {a - c} \right)}^2}} \right)\\
 = \left( {a - c} \right).\left( {{a^2} + 3{b^2} + {c^2} - 3ab + ac - 3bc - {a^2} + 2ac - {c^2}} \right)\\
 = \left( {a - c} \right).\left( {3{b^2} - 3ab + 3ac - 3bc} \right)\\
 = 3\left( {a - c} \right).\left( {b\left( {b - a} \right) - c\left( {b - a} \right)} \right)\\
 = 3\left( {a - c} \right)\left( {b - a} \right)\left( {b - c} \right)\\
b){a^3}\left( {c - b} \right) + {b^3}\left( {a - c} \right) + {c^3}\left( {b - a} \right)\\
 = {a^3}c - {a^3}b + a{b^3} - {b^3}c + {c^3}\left( {b - a} \right)\\
 = c\left( {{a^3} - {b^3}} \right) - ab\left( {{a^2} - {b^2}} \right) - {c^3}\left( {a - b} \right)\\
 = c\left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) - ab\left( {a - b} \right)\left( {a + b} \right) - {c^3}\left( {a - b} \right)\\
 = \left( {a - b} \right)\left( {{a^2}c + abc + {b^2}c - ab\left( {a + b} \right) - {c^3}} \right)\\
 = \left( {a - b} \right)\left( {{a^2}c + abc + {b^2}c - {a^2}b - a{b^2} - {c^3}} \right)\\
 = \left( {a - b} \right)\left( {{a^2}c - {a^2}b + abc - a{b^2} + {b^2}c - {c^3}} \right)\\
 = \left( {a - b} \right)\left( { - {a^2}\left( {b - c} \right) - ab\left( {b - c} \right) + c\left( {{b^2} - {c^2}} \right)} \right)\\
 = \left( {a - b} \right)\left( {b - c} \right)\left( { - {a^2} - ab + c\left( {b + c} \right)} \right)\\
 = \left( {a - b} \right)\left( {b - c} \right).\left( {{c^2} - {a^2} + bc - ab} \right)\\
 = \left( {a - b} \right)\left( {b - c} \right).\left( {c - a} \right)\left( {c + a + b} \right)
\end{array}$