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

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asked Oct 10, 2019 90.2k views

1 Answer

Thecla · Answer 1 · 2019-10-15T18:42:17+0000

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

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

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