Đáp án:
$\begin{array}{l}
{a^3} + {b^3} + {c^3} + 3\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)\\
= {a^3} + {b^3} + {c^3} + \left( {3a + 3b} \right)\left( {bc + ab + {c^2} + ac} \right)\\
= {a^3} + {b^3} + {c^3} + 3abc + 3{a^2}b + 3a{c^2} + 3{a^2}c\\
+ 3{b^2}c + 3a{b^2} + 3b{c^2} + 3abc\\
= {a^3} + 3{a^2}b + 3a{b^2} + {b^3}\\
+ 6abc + 3{a^2}c + 3{b^2}c + 3a{c^2} + 3b{c^2} + {c^3}\\
= {\left( {a + b} \right)^3} + 3\left( {{a^2} + 2ab + {b^2}} \right)c + 3\left( {a + b} \right){c^2} + {c^3}\\
= {\left( {a + b} \right)^3} + 3{\left( {a + b} \right)^2}c + 3\left( {a + b} \right){c^2} + {c^3}\\
= {\left( {a + b + c} \right)^3}\\
\Leftrightarrow {\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 3\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)
\end{array}$
