Đặt \(x^2+x+2=a;x+1=b\Rightarrow a+b=x^2+2x+3\)
PT đã cho trở thành : \(a^3+b^3=\left(a+b\right)^3\)
\(\Leftrightarrow3ab\left(a+b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\b=0\\a+b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+x+2=0\\x+1=0\\x^2+2x+3=0\end{matrix}\right.\)
Do \(x^2+x+2=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}>0\forall x;x^2+2x+3=\left(x+1\right)^2+2>0\forall x\)
\(\Rightarrow x+1=0\Leftrightarrow x=-1\)
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