Với a3 + b3 + c3 = 3abc thì — Không quảng cáo

Phương pháp giải

Sử dụng đẳng thức đặc biệt \({a^3}\; + {b^3}\; + {c^3}\; - 3abc = \;\left( {a + b + c} \right)\left( {{a^2}\; + {b^2}\; + {c^2}\; - ab - bc - ac} \right)\);

Từ đẳng thức đã cho suy ra \({a^3}\; + {b^3}\; + {c^3}\;-3abc = 0\)

\(\begin{array}{*{20}{l}}{{b^3}\; + {c^3}\; = \left( {b + c} \right)\left( {{b^2}\; + {c^2}\;-bc} \right)}\\{ = \left( {b + c} \right)\left[ {{{\left( {b + c} \right)}^2}\;-3bc} \right]}\\{ = {{\left( {b + c} \right)}^3}\;-3bc\left( {b + c} \right)}\\{ \Rightarrow {a^3}\; + {b^3}\; + {c^3}\;-3abc = {a^3}\; + \left( {{b^3}\; + {c^3}} \right)-3abc}\\{ \Leftrightarrow {a^3}\; + {b^3}\; + {c^3}\;-3abc = {a^3}\; + \left( {{b^3}\; + {c^3}} \right)-3abc\left( {b + c} \right)-3abc}\\{ \Leftrightarrow {a^3}\; + \left( {{b^3}\; + {c^3}} \right)-3abc = \left( {a + b + c} \right)\left( {{a^2}\;-a\left( {b + c} \right) + {{\left( {b + c} \right)}^2}} \right)-\left[ {3bc\left( {b + c} \right) + 3abc} \right]}\\{ \Leftrightarrow {a^3}\; + \left( {{b^3}\; + {c^3}} \right)-3abc = \left( {a + b + c} \right)\left( {{a^2}\;-a\left( {b + c} \right) + {{\left( {b + c} \right)}^2}} \right)-3bc\left( {a + b + c} \right)}\\{ \Leftrightarrow {a^3}\; + \left( {{b^3}\; + {c^3}} \right)-3abc = \left( {a + b + c} \right)\left( {{a^2}\;-a\left( {b + c} \right) + {{\left( {b + c} \right)}^2}\;-3bc} \right)}\\{ \Leftrightarrow {a^3}\; + \left( {{b^3}\; + {c^3}} \right)-3abc = \left( {a + b + c} \right)\left( {{a^2}\;-ab\; - ac + {b^2}\; + 2bc + {c^2}\;-3bc} \right)}\\{ \Leftrightarrow {a^3}\; + \left( {{b^3}\; + {c^3}} \right)-3abc = \left( {a + b + c} \right)\left( {{a^2}\; + {b^2}\; + {c^2}\;-ab-ac-bc} \right)}\end{array}\)

Do đó nếu \({a^3}\; + \left( {{b^3}\; + {c^3}} \right)-3abc = 0\) thì \(a + b + c\; = 0\) hoặc \({a^2}\; + {b^2}\; + {c^2}\;-ab-ac-bc = 0\)

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

Nếu \({\left( {a-b} \right)^2}\; + {\left( {a-c} \right)^2}\; + {\left( {b-c} \right)^2}\; = 0 \Leftrightarrow \;\left\{ \begin{array}{l}a - b = 0\\b - c = 0\\a - c = 0\end{array} \right. \Rightarrow a = b = c\)

Vậy \({a^3}\; + \left( {{b^3}\; + {c^3}} \right) = 3abc\) thì \(a = b = c\) hoặc \(a + b + c = 0\).