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since \(3^2=9\) you can forget 1 and 4

im thinkin its 2

\[ (3xy^4)^2(y^2)^3\] square everything first, i.e. multiply all the exponents by 2

9x^2y^11

you distribute the exponents outside of the parentheses to all the terms inside the parentheses

then
\[3^2x^4y^8\times y^6\] now add up the exponents

you get
\[9x^2y^{14}\]

oops i made a mistake above, there is no \(x^4\) term it is \(x^2\)

you are right, it is 2

thanks !!