Range of a

(-1,8)

\hspace{-16}\bf{y=f(x)=x^4+8x^3+18x^2+8x=-a}$\\\\\\ $\bf{\frac{dy}{dx}=4x^3+24x^2+36x+8}$\\\\\\ $\bf{\frac{d^2y}{dx^2}=12x^2+48x+36}$\\\\\\ Now For Max. Or Min. $\bf{\frac{dy}{dx}=0}$\\\\\\ So $\bf{4x^3+24x^2+36x+8=0}$\\\\\\ $\bf{4 \big(x^3+6x^2+9x+2\big) =0}$\\\\\\ $\bf{x=-2\;\;,-2+\sqrt{3}\;\;,-2-\sqrt{3}}$\\\\\\ $\bf{\frac{d^2y}{dx^2}=12\big(x^2+4x+3\big)=12(x+1)(x+3)}$\\\\\\ So $\bf{\frac{d^2y}{dx^2}|_{x=-2}=-}$(ve)\\\\\\ Means $\bf{y=f(x)}$ is Min. at $\bf{x=-2}$\\\\\\ Similarly $\bf{\frac{d^2y}{dx^2}|_{x=-2+\sqrt{3}\;,-2-\sqrt{3}}=(+)}$(ve)\\\\\\

\hspace{-16}$So $\bf{y=f(x)}$ is Min. at $\bf{x=-2+\sqrt{3}\;,-2-\sqrt{3}}$\\\\\\ So $\bf{y=f(x)_{Max}=8}$ and $\bf{y=f(x)_{Min}=-1}$\\\\\\ So $\bf{-1<-a<8\Leftrightarrow a\in \big(-8,1\big)}$