Giải thích các bước giải:

\[\begin{array}{l}
a,\\
{\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)\\
 \Rightarrow {a^3} + {b^3} = {\left( {a + b} \right)^3} - 3ab\left( {a + b} \right)\\
 \Rightarrow {a^3} + {b^3} + {c^3} - 3abc\\
 = \left( {{{\left( {a + b} \right)}^3} - 3ab\left( {a + b} \right)} \right) + {c^3} - 3abc\\
 = \left( {{{\left( {a + b} \right)}^3} + {c^3}} \right) - 3ab\left( {a + b + c} \right)\\
 = {\left( {a + b + c} \right)^3} - 3\left( {a + b} \right)c\left( {a + b + c} \right) - 3ab\left( {a + b + c} \right)\\
 = {\left( {a + b + c} \right)^3} - 3\left( {a + b + c} \right)\left( {ac + bc + ab} \right)\\
 = \left( {a + b + c} \right)\left[ {{{\left( {a + b + c} \right)}^2} - 3\left( {ab + bc + ca} \right)} \right]\\
 = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca - 3ab - 3bc - 3ca} \right)\\
 = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)\\
b,
\end{array}\]

Áp dụng phần a ta có:

\[\begin{array}{l}
{a^3} + {b^3} + {c^3} - 3abc = {\left( {a + b + c} \right)^3} - 3\left( {a + b + c} \right)\left( {ab + bc + ca} \right)\\
 \Leftrightarrow {\left( {a + b + c} \right)^3} - {a^3} - {b^3} - {c^3} = 3\left( {a + b + c} \right)\left( {ab + bc + ca} \right) - 3abc\\
 \Leftrightarrow {\left( {a + b + c} \right)^3} - {a^3} - {b^3} - {c^3} = 3\left( {{a^2}b + abc + {a^2}c + {b^2}a + {b^2}c + abc + abc + {c^2}b + {c^2}a - abc} \right)\\
 \Leftrightarrow {\left( {a + b + c} \right)^3} - {a^3} - {b^3} - {c^3} = 3\left( {ab\left( {a + b} \right) + c\left( {{a^2} + {b^2} + 2ab} \right) + {c^2}\left( {a + b} \right)} \right)\\
 \Leftrightarrow {\left( {a + b + c} \right)^3} - {a^3} - {b^3} - {c^3} = 3\left( {a + b} \right)\left( {ab + ca + cb + {c^2}} \right)\\
 \Leftrightarrow {\left( {a + b + c} \right)^3} - {a^3} - {b^3} - {c^3} = 3\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)
\end{array}\]