Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+1}=a\\\sqrt[3]{x-1}=b\end{matrix}\right.\) \(\Rightarrow\frac{a^3-b^3}{2}=1\)
Pt trở thành:
\(a^2+b^2+ab=\frac{a^3-b^3}{2}\)
\(\Leftrightarrow2\left(a^2+b^2+ab\right)=\left(a-b\right)\left(a^2+b^2+ab\right)\)
\(\Leftrightarrow\left(a^2+b^2+ab\right)\left(a-b-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a^2+b^2+ab=0\Leftrightarrow a=b=0\left(vn\right)\\a-b-2=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt[3]{x+1}-\sqrt[3]{x-1}-2=0\)
Đặt \(\sqrt[3]{x-1}=t\Rightarrow x=t^3+1\)
\(\Rightarrow\sqrt[3]{t^3+2}=t+2\)
\(\Leftrightarrow t^3+2=t^3+6t^2+12t+8\)
\(\Leftrightarrow6\left(t+1\right)^2=0\Rightarrow t=-1\)
\(\Rightarrow\sqrt[3]{x-1}=-1\Rightarrow x=0\)
