Với $a^3 + b^3 + c^3$ = 3abc thì 

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 = $\frac{1}{2}$[${(a – b)}^2 + {(a – c)}^2 + {(b – c)}^2$]

Nếu ${(a – b)}^2 + {(a – c)}^2 + {(b – c)}^2$ = 0 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