Ta có: \(\left(x+2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3+6x^2+12x+8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3+27+6x^2+12x+6-49=0\)
\(\Leftrightarrow12x^2+24x-8=0\)
\(\Leftrightarrow12\left(x^2+2x-\frac{2}{3}\right)=0\)
\(\Leftrightarrow x^2+2x+1-\frac{5}{3}=0\)
\(\Leftrightarrow\left(x+1\right)^2=\frac{5}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\frac{\sqrt{5}}{\sqrt{3}}\\x+1=-\frac{\sqrt{5}}{\sqrt{3}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{15}-\sqrt{3}}{3}\\x=\frac{-\sqrt{15}-\sqrt{3}}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{\sqrt{15}-\sqrt{3}}{3};\frac{-\sqrt{15}-\sqrt{3}}{3}\right\}\)
