Ta có:
$\left(\dfrac 1a +\dfrac 1b + \dfrac 1c\right)^2 = \dfrac{1}{a^2} +\dfrac{1}{b^2} + \dfrac{1}{c^2}$
$\Leftrightarrow \dfrac{1}{ab} + \dfrac{1}{bc} +\dfrac{1}{ca}=0$
$\Leftrightarrow a + b + c = 0$
Do đó:
$a^3 + b^3 + c^3$
$= a^3 + b^3 + c^3 - (a + b + c)$
$= a^3 - a + b^3 - b + c^3 - c$
$= (a-1)a(a + 1) + (b-1)b(b +1) + (c-1)c(c+1)$
Do $\begin{cases}(a-1)a(a + 1)\quad \vdots \quad 6\\(b-1)b(b + 1)\quad \vdots \quad 6\\(b-1)b(b + 1)\quad \vdots \quad 6\end{cases}$
nên $(a-1)a(a + 1) + (b-1)b(b +1) + (c-1)c(c+1) \quad \vdots \quad 6$
Vậy $a^3 + b^3 + c^3\quad \vdots \quad 6$
