Đặt \(x+4=t\) pt trở thành:
\(\sqrt[3]{t-2}+\sqrt[3]{t}+\sqrt[3]{t+2}=0\Leftrightarrow\sqrt[3]{t-2}+\sqrt[3]{t+2}=-\sqrt[3]{t}\)
\(\Leftrightarrow t-2+3\sqrt[3]{t^2-4}\left(\sqrt[3]{t-2}+\sqrt[3]{t+2}\right)+t+2=-t\)
\(\Leftrightarrow3\sqrt[3]{t^2-4}.\left(-\sqrt[3]{t}\right)=-3t\)
\(\Leftrightarrow\sqrt[3]{t^3-4t}=t\Leftrightarrow t^3-4t=t^3\)
\(\Leftrightarrow-4t=0\Rightarrow t=0\Rightarrow x+4=0\Rightarrow x=-4\)
Vậy pt có nghiệm duy nhất \(x=-4\)