\(D=\frac{\left(x^2\right)^3+8^3}{x^2+8}=\frac{\left(x^2+8\right)\left(x^4-8x^2+64\right)}{x^2+8}\)
\(=x^4-8x^2+64=\left(x^2-4\right)^2+48\ge48\left(\forall x\right)\)
Dấu "=" xảy ra khi \(x^2-4=0\Leftrightarrow x=\pm2\)
Vậy \(D_{min}=48\Leftrightarrow x=\pm2\)