Chứng minh rằng với mọi \(a,b,c\) ta luôn có:

\({\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 3\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right).\)

Ta có: \({\left( {a + b + c} \right)^3} = {\left( {a + b} \right)^3} + 3{\left( {a + b} \right)^2}c + 3\left( {a + b} \right){c^2} + {c^3}\)

\( = {a^3} + 3{a^2}b + 3a{b^2} + {b^3} + 3{\left( {a + b} \right)^2}c + 3\left( {a + b} \right){c^2} + {c^3}\)

\( = {a^3} + {b^3} + {c^3} + 3ab\left( {a + b} \right) + 3{\left( {a + b} \right)^2}c + 3\left( {a + b} \right){c^2}\)

\( = {a^3} + {b^3} + {c^3} + 3\left( {a + b} \right)\left[ {ab + \left( {a + b} \right)c + {c^2}} \right]\)

\[ = {a^3} + {b^3} + {c^3} + 3\left( {a + b} \right)\left( {ab + ac + bc + {c^2}} \right)\]

\[ = {a^3} + {b^3} + {c^3} + 3\left( {a + b} \right)\left[ {a\left( {b + c} \right) + c\left( {b + c} \right)} \right]\]

\[ = {a^3} + {b^3} + {c^3} + 3\left( {a + b} \right)\left( {b + c} \right)\left( {a + c} \right)\].