Rút gọn biểu thức:(a+b)^3+(b+c)^3+(c+a)^3-3(a+b)(b+c)(c+a)

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

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