Đáp án:
Giải thích các bước giải:
$\begin{array}{l}
{\left( {a + b + c} \right)^3} - {a^3} - {b^3} - {c^3}\\
= \left[ {{{\left( {a + b + c} \right)}^3} - {a^3}} \right] - \left( {{b^3} + {c^3}} \right)\\
= \left\{ {\left( {a + b + c - a} \right)\left[ {{{\left( {a + b + c} \right)}^2} + a\left( {a + b + c} \right) + {a^2}} \right]} \right\} - \left( {b + c} \right)\left( {{b^2} - bc + {c^2}} \right)\\
= \left( {b + c} \right)\left[ {{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca + {a^2} + ab + ac + {a^2}} \right] - \left( {b + c} \right)\left( {{b^2} - bc + {c^2}} \right)\\
= \left( {b + c} \right)\left[ {3{a^2} + {b^2} + {c^2} + 3ab + 2bc + 3ac - {b^2} + bc - {c^2}} \right]\\
= \left( {b + c} \right)\left( {3{a^2} + 3ab + 3ac + 3bc} \right)\\
= 3\left( {b + c} \right)\left( {{a^2} + ab + ac + bc} \right)\\
= 3\left( {b + c} \right)\left[ {a\left( {a + b} \right) + c\left( {a + b} \right)} \right]\\
= 3\left( {b + c} \right)\left( {a + b} \right)\left( {a + c} \right)
\end{array}$
