Ta có: \(\left(x+1\right)^3-x\left(x-2\right)\left(x+2\right)-3x^2=-2\)
\(\Leftrightarrow x^3+3x^2+3x+1-x\left(x^2-4\right)-3x^2=-2\)
\(\Leftrightarrow x^3+3x+1-x^3+4x=-2\)
\(\Leftrightarrow7x+1=-2\)
\(\Leftrightarrow7x=-3\Leftrightarrow x=-\dfrac{3}{7}\)
Vậy x = \(-\dfrac{3}{7}\)