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Which value of b solves the equation? b^-3= = 1/216

Sagot :

[tex]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}[/tex]


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