<=>\(\sqrt[3]{x+1}=x+1\)
Lập phương 2 vế ta được
\(x+1=\left(x+1\right)^3\Leftrightarrow x+1=x^3+3x^2+3x+1\)
\(\Leftrightarrow x^3+3x^2+2x=0\Leftrightarrow x\left(x^2+3x+2\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
Vậy S={0;-1;-2}
\(\sqrt[3]{x+1}-x=1\)
\(\Leftrightarrow x+1=\left(x+1\right)^3\)
\(\Leftrightarrow x^3+3x^2+3x+1-x-1=0\)
\(\Leftrightarrow x^3+3x^2+2x=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+x\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
Vậy tập nghiệm của PT là : S = {0;-1;-2}
