Đặt: \(\hept{\begin{cases}\sqrt[3]{x}=a\\\sqrt[3]{y}=b\\\sqrt[3]{z}=c\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y+z=a^3+b^3+c^3\\3\sqrt[3]{xyz}=3abc\end{cases}}\) Theo hđt mở rộng: \(a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)+3abc\)
\(=\left(\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}\right)\left(a^2+b^2+c^2-ab-ac-bc\right)+3abc=3abc\)
Vậy \(a^3+b^3+c^3=3abc\Leftrightarrow x+y+z=3\sqrt[3]{xyz}\)
Ta có : \(\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}=0\)
\(\Leftrightarrow\sqrt[3]{x}+\sqrt[3]{y}=-\sqrt[3]{z}\)
\(\Leftrightarrow\left(\sqrt[3]{x}+\sqrt[3]{y}\right)^3=-z\)(1)
\(\Leftrightarrow x+y+3\sqrt[3]{x}^2\sqrt[3]{y}+3\sqrt[3]{x}.\sqrt[3]{y}^2=-z\)
\(\Leftrightarrow x+y+3\sqrt[3]{x}\sqrt[3]{y}\left(\sqrt[3]{x}+\sqrt[3]{y}\right)=-z\)
\(\Leftrightarrow x+y-3\sqrt[3]{x}\sqrt[3]{y}\sqrt[3]{z}=-z\left(theo\left(1\right)\right)\)
\(\Leftrightarrow x+y-3\sqrt[3]{xyz}=-z\)
\(\Leftrightarrow x+y+z=3\sqrt[3]{xyz}\)
