Which value of b solves the equation? b^-3= = 1/216
\(b^{-3}=\frac{1}{216} \\\\ \frac{1}{b^3}= \frac{1}{216} \\\\ \frac{1}{b^3}=\frac{1}{6^3} \\\\ b^3=6^3 \\\\ \boxed{b=6}\)
if b^-3=1/216 find the reciprocal (^-1) of each side so it will be b^3=216 and then find the cube root so b=∛216 which is 6
