Cho $a^3 + b^3 + c^3$ = 3abc thì

Từ đẳng thức đã cho suy nghĩ $a^3 + b^3 + c^3$ – 3abc = 0

$$\begin{gathered} B^3 + c^3 = (b + c)(b^2 + c^2 – bc) \\ = (b + c){[{(b + c)}^2 – 3bc]}^4 \\ = {(b + c)}^3 – 3bc(b + c) \end{gathered}$$

$$\begin{gathered} => a^3 + b^3 + c^3 – 3abc \\ = a^3 + (b^3 + c^3) – 3abc \\ \iff a^3 + b^3 + c^3 – 3abc \\ = a^3 + (b^3 + c^3) – 3bc(b + c) – 3abc \\ \iff a^3 + b^3 + c^3 – 3abc \\ = (a + b + c)(a^2 – a(b + c) + {(b \\ + c)}^2) – \left[3bc\right(b + c) + 3abc] \\ \iff a^3 + b^3 + c^3 – 3abc \\ = (a + b + c)(a^2 – a(b + c) + {(b + c)}^2) – 3bc) \\ \iff a^3 + b^3 + c^3 – 3abc \\ = (a + b + c\left)\right(a^2 – ab – ac + b^2 + 2bc + c^2 – 3bc) \\ \iff a^3 + b^3 + c^3 – 3abc \\ = (a + b + c\left)\right(a^2 + b^2 + c^2 – ab – ac – bc) \end{gathered}$$

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 = .[${(a – b)}^2 + {(a – c)}^2 + {(b – c)}^2$]

Suy ra a = b = c

Đáp án cần chọn là: B