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21 January, 16:58

How do I evaluate (27x^3/8y^9) ^-5/3

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  1. 21 January, 17:22
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    First calculate 27 ^ - 5/3 = 1 / 27^5/3 = 1 / 243,

    x^3 ^ - 5/3 = 1 / x^3 ^5/3 = 1 / x^5

    8 ^-5/3 = 1 / 8^5/3 = 1/32

    y^9 ^ - 5/3 = 1 / y^9^5/3 = 1 / y^15

    so we have 1/243 x^5 * 32 * y^15

    = 32y^15 / 243x^5 Answer
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