We want to evaluate
[tex]b^2c^{-1}[/tex]for b = 8 and c = -4
Solution
From
[tex]\begin{gathered} b^2c^{-1}=b^2\times\frac{1}{c} \\ b^2c^{-1}=\frac{b^2}{c} \end{gathered}[/tex]Put b = 8 and c = -4
[tex]\begin{gathered} b^2c^{-1}=\frac{b^2}{c} \\ b^2c^{-1}=\frac{8^2}{-4} \\ b^2c^{-1}=\frac{64}{-4} \\ b^2c^{-1}=-16 \end{gathered}[/tex]Therefore,
[tex]b^2c^{-1}=-16[/tex]