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Mathematics 16 Online
OpenStudy (anonymous):

Simplify (y3)2.

OpenStudy (phoenixfire):

\[\large (y^3)^2\] That what you mean?

OpenStudy (anonymous):

y^6

OpenStudy (anonymous):

yes @PhoenixFire

OpenStudy (phoenixfire):

\[\large (y^a)^b=y^{ab}\] So, (2)(3)=6 \[y^6\]

OpenStudy (anonymous):

oops no sorry @PhoenixFire [(1/3y^3)^2\]

OpenStudy (phoenixfire):

\[\large ({1 \over 3}y^3)^2\] or \[\large ({1 \over {3y^3}})^2\]

OpenStudy (anonymous):

first one sorry i was so unclear im new to this haha @PhoenixFire

OpenStudy (phoenixfire):

It's alright. For the first one the rule still applies partially \[\large ({1 \over 3}y^3)^2=({1 \over 3})^2(y^3)^2=({1^2 \over 3^2})(y^{(3)(2)})\] You can simplify that.

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