\(\Leftrightarrow2x^3+6x^2+6x+2-5x-1-\sqrt[3]{3x+1}=0\)
\(\Leftrightarrow2\left(x+1\right)^3+x+1-2\left(3x+1\right)-\sqrt[3]{3x+1}=0\)
Đặt \(\left\{{}\begin{matrix}x+1=a\\\sqrt[3]{3x+1}=b\end{matrix}\right.\) phương trình trở thành:
\(2a^3+a-2b^3-b=0\)
\(\Leftrightarrow2\left(a-b\right)\left(a^2+ab+b^2\right)+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a^2+2ab+2b^2+1\right)=0\)
\(\Leftrightarrow a=b\) (do \(2\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{2}+1>0\))
\(\Leftrightarrow x+1=\sqrt[3]{3x+1}\)
\(\Leftrightarrow x^3+3x^2+3x+1=3x+1\)
\(\Leftrightarrow x^3+3x^2=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
