\(\begin{array}{l}
P = {\left( {a + b} \right)^3} + {\left( {b + c} \right)^3} + {\left( {c + a} \right)^3} - 3\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)\\
= {a^3} + 3ab\left( {a + b} \right) + {b^3} + {b^3} + 3bc\left( {b + c} \right) + {c^3} + {c^3} + 3ca\left( {a + c} \right) + {a^3} - 3\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)\\
= 2{a^3} + 2{b^3} + 2{c^3} + 3ab\left( {a + b} \right) + 3bc\left( {b + c} \right) + 3ca\left( {a + c} \right) - 3a\left( {b + c} \right)\left( {c + a} \right) - 3b\left( {b + c} \right)\left( {c + a} \right)\\
= 2{a^3} + 2{b^3} + 2{c^3} + 3ab\left( {a + b} \right) + \left( {b + c} \right)\left[ {3bc - 3b\left( {c + a} \right)} \right] + \left( {a + c} \right)\left[ {3ac - 3a\left( {b + c} \right)} \right]\\
= 2{a^3} + 2{b^3} + 2{c^3} + 3ab\left( {a + b} \right) + \left( {b + c} \right)\left( {3bc - 3bc - 3ba} \right) + \left( {a + c} \right)\left( {3ac - 3ab - 3ac} \right)\\
= 2{a^3} + 2{b^3} + 2{c^3} + 3{a^2}b + 3a{b^2} - 3ab\left( {b + c} \right) - 3ab\left( {a + c} \right)\\
= 2{a^3} + 2{b^3} + 2{c^3} + 3{a^2}b + 3a{b^2} - 3a{b^2} - 3abc - 3{a^2}b - 3abc\\
= 2{a^3} + 2{b^3} + 2{c^3} - 6abc.
\end{array}\)
