ĐKXĐ: \(x\ge-\frac{1}{3}\)
\(\Leftrightarrow\left(\sqrt{3x+1}\right)^3=\left(x^2+1\right)^3\)
\(\Leftrightarrow\sqrt{3x+1}=x^2+1\)
\(\Leftrightarrow3x+1=x^4+2x^2+1\)
\(\Leftrightarrow x^4+2x^2-3x=0\)
\(\Leftrightarrow x\left(x^3+2x-3\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x^2+x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)