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monitta
3 years ago
14

How to solve 8 to the power of negative one third

Mathematics
1 answer:
wariber [46]3 years ago
8 0

8^{-\frac{1}{3}}=\dfrac{1}{8^\frac{1}{3}}=\dfrac{1}{\sqrt[3]8}=\dfrac{1}{2}\\\\Used:\\\\a^{-n}=\dfrac{1}{a^n}\\\\a^\frac{1}{n}=\sqrt[n]{a}\\------------------------\\Other\ method\\\\8=2^3\\\\8^{-\frac{1}{3}}=(2^3)^{-\frac{1}{3}}=2^{3\cdot\left(-\frac{1}{3}\right)}=2^{-1}=\dfrac{1}{2}\\\\Used:\\\\(a^n)^m=a^{n\cdot m}\\\\a^{-n}=\dfrac{1}{a^n}

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