Đáp án:
$\begin{array}{l}
a)\left( {x + a} \right)\left( {x + b} \right)\left( {x + c} \right)\\
= \left( {{x^2} + bx + ax + ab} \right)\left( {x + c} \right)\\
= {x^3} + c{x^2} + b{x^2} + bcx + a{x^2} + acx + abx + abc\\
= {x^3} + \left( {a + b + c} \right){x^2} + \left( {ab + bc + ac} \right)x + abc\\
\Leftrightarrow \left( {x + a} \right)\left( {x + b} \right)\left( {x + c} \right)\\
= {x^3} + \left( {a + b + c} \right){x^2} + \left( {ab + bc + ac} \right)x + abc\\
b){a^3} + {b^3} + {c^3} - 3abc\\
= {a^3} + 3{a^2}b + 3a{b^2} + {b^3} + {c^3} - 3{a^2}b - 3a{b^2} - 3abc\\
= {\left( {a + b} \right)^3} + {c^3} - 3ab\left( {a + b + c} \right)\\
= \left( {a + b + c} \right)\left( {{{\left( {a + b} \right)}^2} - c\left( {a + b} \right) + {c^2}} \right) - 3ab\left( {a + b + c} \right)\\
= \left( {a + b + c} \right)\left( {{a^2} + 2ab + {b^2} - ac - bc + {c^2} - 3ab} \right)\\
= \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - ac - bc} \right)\\
\Leftrightarrow \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - ac - bc} \right)\\
= {a^3} + {b^3} + {c^3} - 3abc\\
c)\left( {a - b} \right)\left( {b - c} \right)\left( {a - c} \right)\\
= \left( {ab - ac - {b^2} + bc} \right)\left( {a - c} \right)\\
= {a^2}b - {a^2}c - a{b^2} + abc - abc + a{c^2} + {b^2}c - b{c^2}\\
= {a^2}b - {a^2}c - a{b^2} + {b^2}c + a{c^2} - b{c^2}\\
= {a^2}\left( {b - c} \right) + {b^2}\left( {c - a} \right) + {c^2}\left( {a - b} \right)\\
\Leftrightarrow {a^2}\left( {b - c} \right) + {b^2}\left( {c - a} \right) + {c^2}\left( {a - b} \right)\\
= \left( {a - b} \right)\left( {b - c} \right)\left( {a - c} \right)
\end{array}$
