Ta có: $\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0$

$\Rightarrow \dfrac{yz + zx + xy}{xyz} = 0$

$\Rightarrow xy + yz + zx = 0$

Đặt $A = \dfrac{1}{x}$, $B = \dfrac{1}{y}$, $C = \dfrac{1}{z}$

$\Rightarrow A + B + C = 0$

Mà: $A^3 + B^3 + C^3 - 3ABC = (A + B + C)(A^2 + B^2 + C^2 - AB - BC - CA)$

$\Rightarrow A^3 + B^3 + C^3 - 3ABC = 0 \cdot (A^2 + B^2 + C^2 - AB - BC - CA)$

$\Rightarrow A^3 + B^3 + C^3 - 3ABC = 0$

$\Rightarrow A^3 + B^3 + C^3 = 3ABC$

hay $\dfrac{1}{x^3} + \dfrac{1}{y^3} + \dfrac{1}{z^3} = \dfrac{3}{xyz}$

Xét vế trái

$= \dfrac{1}{x^3} + \dfrac{1}{y^3} + \dfrac{1}{z^3}$

$= \dfrac{yz}{x^2} + \dfrac{zx}{y^2} + \dfrac{xy}{z^2}$

$= \dfrac{xyz}{x^3} + \dfrac{xyz}{y^3} + \dfrac{xyz}{z^3}$

$= xyz \cdot \left(\dfrac{1}{x^3} + \dfrac{1}{y^3} + \dfrac{1}{z^3}\right)$

$= xyz \cdot \dfrac{3}{xyz} = 3 = $ Vế phải (đpcm)

$\color{#FF0000}{♡^♡}\color{#FF7F00}{𝖘}\color{#FFD700}{𝖒}\color{#00FF00}{𝖕}\color{#00BFFF}{9}\color{#0000FF}{9}\color{#9400D3}{9}\color{#FF0000}{♡^♡}$