Phương pháp giải:
Sử dụng đẳng thức đặc biệt a 3 + b 3 + c 3 − 3 a b c = ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − a c ) {a^3}\; + {b^3}\; + {c^3}\; - 3abc = \;\left( {a + b + c} \right)\left( {{a^2}\; + {b^2}\; + {c^2}\; - ab - bc - ac} \right) a 3 + b 3 + c 3 − 3 ab c = ( a + b + c ) ( a 2 + b 2 + c 2 − ab − b c − a c )
Lời giải chi tiết:
Từ đẳng thức đã cho suy ra a 3 + b 3 + c 3 − 3 a b c = 0 {a^3}\; + {b^3}\; + {c^3}\; - 3abc = 0 a 3 + b 3 + c 3 − 3 ab c = 0
b 3 + c 3 = ( b + c ) ( b 2 + c 2 − b c ) {b^3}\; + {c^3}\; = \left( {b + c} \right)\left( {{b^2}\; + {c^2}\; - bc} \right) b 3 + c 3 = ( b + c ) ( b 2 + c 2 − b c ) = ( b + c ) [ ( b + c ) 2 − 3 b c ] = \left( {b + c} \right)\left[ {{{\left( {b + c} \right)}^2}\; - 3bc} \right] = ( b + c ) [ ( b + c ) 2 − 3 b c ] = ( b + c ) 3 − 3 b c ( b + c ) = {\left( {b + c} \right)^3}\; - 3bc\left( {b + c} \right) = ( b + c ) 3 − 3 b c ( b + c ) ⇒ a 3 + b 3 + c 3 − 3 a b c = a 3 + ( b 3 + c 3 ) − 3 a b c \Rightarrow {a^3}\; + {b^3}\; + {c^3}\; - 3abc = {a^3}\; + \left( {{b^3}\; + {c^3}} \right) - 3abc ⇒ a 3 + b 3 + c 3 − 3 ab c = a 3 + ( b 3 + c 3 ) − 3 ab c = a 3 + ( b + c ) 3 − 3 b c ( b + c ) − 3 a b c = {a^3}\; + {\left( {b + c} \right)^3} - 3bc\left( {b + c} \right) - 3abc = a 3 + ( b + c ) 3 − 3 b c ( b + c ) − 3 ab c = ( a + b + c ) ( a 2 − a ( b + c ) + ( b + c ) 2 ) − [ 3 b c ( b + c ) + 3 a b c ] = \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] = ( a + b + c ) ( a 2 − a ( b + c ) + ( b + c ) 2 ) − [ 3 b c ( b + c ) + 3 ab c ] = ( a + b + c ) ( a 2 − a ( b + c ) + ( b + c ) 2 ) − 3 b c ( a + b + c ) = \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) = ( a + b + c ) ( a 2 − a ( b + c ) + ( b + c ) 2 ) − 3 b c ( a + b + c ) = ( a + b + c ) ( a 2 − a ( b + c ) + ( b + c ) 2 − 3 b c ) = \left( {a + b + c} \right)\left( {{a^2}\; - a\left( {b + c} \right) + {{\left( {b + c} \right)}^2}\; - 3bc} \right) = ( a + b + c ) ( a 2 − a ( b + c ) + ( b + c ) 2 − 3 b c ) = ( a + b + c ) ( a 2 − a b − a c + b 2 + 2 b c + c 2 − 3 b c ) = \left( {a + b + c} \right)\left( {{a^2}\; - ab\; - ac + {b^2}\; + 2bc + {c^2}\; - 3bc} \right) = ( a + b + c ) ( a 2 − ab − a c + b 2 + 2 b c + c 2 − 3 b c ) = ( a + b + c ) ( a 2 + b 2 + c 2 − a b − a c − b c ) = \left( {a + b + c} \right)\left( {{a^2}\; + {b^2}\; + {c^2}\; - ab - ac - bc} \right) = ( a + b + c ) ( a 2 + b 2 + c 2 − ab − a c − b c )
Do đó nếu a 3 + ( b 3 + c 3 ) − 3 a b c = 0 {a^3}\; + \left( {{b^3}\; + {c^3}} \right) - 3abc = 0 a 3 + ( b 3 + c 3 ) − 3 ab c = 0 a + b + c = 0 a + b + c\; = 0 a + b + c = 0 a 2 + b 2 + c 2 − a b − a c − b c = 0 {a^2}\; + {b^2}\; + {c^2}\; - ab - ac - bc = 0 a 2 + b 2 + c 2 − ab − a c − b c = 0
Mà a 2 + b 2 + c 2 − a b − a c − b c = [ ( a − b ) 2 + ( a − c ) 2 + ( b − c ) 2 ] {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] a 2 + b 2 + c 2 − ab − a c − b c = [ ( a − b ) 2 + ( a − c ) 2 + ( b − c ) 2 ]
Nếu ( a − b ) 2 + ( a − c ) 2 + ( b − c ) 2 = 0 ⇔ { a − b = 0 b − c = 0 a − c = 0 ⇒ a = b = c {\left( {a - b} \right)^2}\; + {\left( {a - c} \right)^2}\; + {\left( {b - c} \right)^2}\; = 0 \Leftrightarrow \;\left\{ {\begin{array}{llllllllllllllllllll}{a - b = 0}\\{b - c = 0}\\{a - c = 0}\end{array}} \right. \Rightarrow a = b = c ( a − b ) 2 + ( a − c ) 2 + ( b − c ) 2 = 0 ⇔ ⎩ ⎨ ⎧ a − b = 0 b − c = 0 a − c = 0 ⇒ a = b = c
Vậy a 3 + ( b 3 + c 3 ) = 3 a b c {a^3}\; + \left( {{b^3}\; + {c^3}} \right) = 3abc a 3 + ( b 3 + c 3 ) = 3 ab c a = b = c a = b = c a = b = c a + b + c = 0 a + b + c = 0 a + b + c = 0
