\(2\sqrt[3]{\left(x+2\right)^2}-\sqrt[3]{\left(x-2\right)^2}=\sqrt[3]{x^2-4}\)
\(\Leftrightarrow\sqrt[3]{\left(x-2\right)^2}=\sqrt[3]{x^2-4}\)
\(\Rightarrow\left(x-2\right)^2=x^2-4\)
\(\Leftrightarrow x^2-4x+4-x^2+4=0\)
\(\Leftrightarrow-4x+8=0\)
\(\Leftrightarrow x=2\)
Đặt \(\sqrt[3]{x+2}=a;\sqrt[3]{x-2}=b;\) ta có:
\(2a^2-b^2=ab\) ⇔ \(2a^2-ab-b^2=0\)
\(\Leftrightarrow2a^2+ab-2ab-b^2=0\)
⇔ \(\left(2a+b\right)\left(a-b\right)=0\)
⇔ \(\left[{}\begin{matrix}2\sqrt[3]{x+2}=-\sqrt[3]{x-2}\\\sqrt[3]{x-2}=\sqrt[3]{x+2}\end{matrix}\right.\)⇔ \(x=-\frac{14}{9}\)