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In $(3xy)^2$, do I distribute that power of two to each of the terms?

$(3^2)\times(x^2)\times(y^2) = 9x^2y^2$?

Or do I just treat it as $3xy^2$?

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    $\begingroup$ The first option. $\endgroup$ – Edward Jiang Dec 19 '14 at 2:28
  • $\begingroup$ $ 3^2 \cdot x^2 \cdot y^2 \; = \; 9x^2y^2 $ $\endgroup$ – mathamphetamines Dec 19 '14 at 2:29
  • $\begingroup$ you have computer so search it up. on google $\endgroup$ – Irrational Person Dec 19 '14 at 2:31
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    $\begingroup$ I am searching it up, right here. ._. $\endgroup$ – John Dec 19 '14 at 2:32
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$$(3xy)^2 = (3xy)(3xy) = 3\cdot 3\cdot x\cdot x\cdot y\cdot y = 3^2x^2y^2=9x^2y^2$$

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  • $\begingroup$ you got it first $\endgroup$ – Irrational Person Dec 19 '14 at 2:32
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All terms. $(3xy)^2=(3xy)(3xy)=(9x^2y^2)$

Good time to note how it's important to put those parentheses around expressions like this to distinguish $(3xy)^2$ from $3xy^2$

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