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\[(\frac{ 2yx^2*-y^-1z^0 }{ x^-3*2y^0z^-1 })~~ ^3\]

Is this the problem?
\[\Large \left(\frac{2yx^2*-y^{-1}z^0}{x^{-3}*2y^0z^{-1}}\right)^3\]

Yes

I'm not getting the same answer you got

I just got all mixed up, i know how to solve these but i saw this and my mind went blank

y^0 or y^-1 I don't know if i should add them or multiply

add
\[\Large y^1*y^{-1} = y^{1+(-1)} = y^0\]

anything to the zeroth power is 1, so y^0 = 1

ok so that elliminates y

yes so y is nowhere in the final answer

sorry if i have any late replies, i am making sure i right this down

that's fine

I also forgot about the y^0 down below, that also turns into 1 but I'm sure you see that by now

yes, i got that part :)

ok great

ok right now i have x^6 / x^-9 z^-3 ? am i close

now simplify \[\Large \frac{x^6}{x^{-9}z^{-3}}\]

x^15 z^3 ?

correct

the online solver somehow got -x^6 z^3 / x^-9 , is my simplified answer still correct?

well the solver didn't fully simplify

the x^6 over x^(-9) simplifies to x^15

i used cymath.com (because it shows me steps) and ohhhh okay

you're welcome, you did really good on it