\[\sqrt[8]{b^{2} + b - 12}\ \]
\[b^{2} + b - 12 \geq 0\]
\[b_{1} + b_{2} = - 1;\ \ \ b_{1} \cdot b_{2} = - 12\]
\[b_{1} = - 4;\ \ b_{2} = 3\]
\[(b + 4)(b - 3) \geq 0\]
\[b \leq - 4;\ \ b \geq 3.\]
\[Ответ:при\ b \leq - 4;\ \ b \geq 3.\]