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Question
Simplify and express as positive indice;
`[(125"a"^-3)/"y"^6]^(-1/3)`
Sum
Solution
`[(125"a"^-3)/"y"^6]^(-1/3)=[(5^3"a"^-3)/"y"^6]^(-1/3)`
`=(5^(3xx(-1)/3)."a"^(-3xx(-1)/3))/"y"^(6xx(-1)/3)`
`=(5^-1."a"^1)/"y"^-2`
`=(a.y^2)/5`
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Law of Exponents (For Integral Powers)
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