simplify : (-xy^3)^2 x^-3
\[(-xy^3)^2 \cdot x^{-3}\] distribute the power of two to all variables inside the parenthesis first.
(-x^2y^6)x^-3
\[[(-x)^2(y^3)^2] \cdot x^{-3}\] Good.:)
I'm not sure what to do next.
\[(-x)^2 = (-x)(-x) = x\]NOw you've got \[(x\cdot y^6) \cdot x^{-3}\]
Now we multiply \(x^{-3}\) into the stuff with parenthesis.
\[(x \cdot x^{-3}) \cdot (y^6 \cdot x^{-3})\]
\[(x \cdot x^{-3}) = x^{1-3} = (x^{-2})\cdot x^{-3}y^6\]
(x^-2)(y^6 x x^-3)
y^6/x
Actually, let's try it another way.
what would you do after that? multiply the x^-2 times the x^-3?
ok
Say we want to make all the variables positive first. since we have \(x^{-3}\) we can rewrite that as a positive power by putting it over 1. \[\frac{1}{x^3}\]
ok
Now that makes all our variables positive. you could have continued doing it the way we were doing it at first, and still ended up with the same result by making all the negative variables positive at the end, I just thought this second approach might be a bit easier to understand.
so then (-x^2) would be x^2
x^3-x^2
y6/x
So now we have \((-xy^3)\) and \(\frac{1}{x^3}\)
oh yeah, you got it.Good job!
THANK YOU SO MUCH FOR YOUR HELP!
Haha no problem.
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