Từ đẳng thức đã cho suy ra a 3 + b 3 + c 3 – 3abc = 0
b 3 + c 3 = (b + c)( b 2 + c 2 – bc)
= (b + c)[ ( b + c ) 2 – 3bc]
= ( b + c ) 3 – 3bc(b + c)
=> a 3 + b 3 + c 3 – 3abc = a 3 + ( b 3 + c 3 ) – 3abc
ó a 3 + b 3 + c 3 – 3abc = a 3 + ( b + c ) 3 – 3bc(b + c) – 3abc
ó a 3 + ( b 3 + c 3 ) – 3abc = (a + b + c)( a 2 – a ( b + c ) + ( b + c ) 2 ) – [3bc(b + c) + 3abc]
ó a 3 + ( b 3 + c 3 ) – 3abc = (a + b + c)( a 2 – a ( b + c ) + ( b + c ) 2 ) – 3bc(a + b + c)
ó a 3 + ( b 3 + c 3 ) – 3abc = (a + b + c)( a 2 – a ( b + c ) + ( b + c ) 2 – 3bc)
ó a 3 + ( b 3 + c 3 ) – 3abc = (a + b + c)( a 2 – ab - ac + b 2 + 2bc + c 2 – 3bc)
ó a 3 + ( b 3 + c 3 ) – 3abc = (a + b + c)( a 2 + b 2 + c 2 – ab – ac – bc)
Do đó nếu a 3 + ( b 3 + c 3 ) – 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 = .[ ( a – b ) 2 + ( a – c ) 2 + ( b – c ) 2 ]
Nếu ( a – b ) 2 + ( a – c ) 2 + ( b – c ) 2 = 0 ó suy ra a = b = c
Vậy a 3 + ( b 3 + c 3 ) = 3abc thì a = b = c hoặc a + b + c = 0
Đáp án cần chọn là: C
