(3y^7)^3 3^3*y^(7*3)
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9y^9 9*y^9
(3y^7)^3 3^3*y^(7*3) 27*y^21 (mult. together the exponents 3 and 7)
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9y^9 9*y^9 9*y^9
That 27/9 reduces to 3. y^(21) / y^9 reduces to y^12.
Thus, we obtain 3*y^12 (answer)
Please do the next one, asking questions if need be, and showing your work.