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

\[{a^2}\; + {\rm{ }}2ab + {b^2} + 2bc + 2ac + {c^2} = 3ab + 3bc + 3ac\]

\[{a^2}\; + \;{b^2} + {c^2} - ab - bc - ac = 0\]

\[2{a^2}\; + \;2{b^2} + 2{c^2} - 2ab - 2bc - 2ac = 0\]

\[\left( {{a^2}--2ab + {b^2}} \right) + \left( {{b^2}--2bc + {c^2}} \right) + \left( {{c^2}--2ac + {a^2}} \right) = 0\]

\[{\left( {a--b} \right)^2} + {\left( {b--c} \right)^2} + {\left( {c--a} \right)^2} = 0\].

Ta có \[{\left( {a--b} \right)^2} \ge 0\,;\,\,{\left( {b--c} \right)^2} \ge 0\,;\,\,{\left( {c--a} \right)^2} \ge 0\].

Để \[{\left( {a--b} \right)^2} + {\left( {b--c} \right)^2} + {\left( {c--a} \right)^2} = 0\] thì

\(\left\{ \begin{array}{l}{\left( {a--b} \right)^2} = 0\\{\left( {b--c} \right)^2} = 0\\{\left( {c--a} \right)^2} = 0\end{array} \right.\) nên \[\left\{ \begin{array}{l}a--b = 0\\b--c = 0\\c--a = 0\end{array} \right.\] suy ra \[\left\{ \begin{array}{l}a = b\\b = c\\c = a\end{array} \right.\] hay \(a = b = c\).

Do đó, nếu \[{\left( {a + b + c} \right)^2} = 3\left( {ab + bc + ac} \right)\] thì \[a = b = c\] (đpcm).

Đa thức 3x^3 y + x^5 - 4x^2 y + 6 có bậc là