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Solve (2x-3y^4)^3 (x^3 + y)^0 divide (4xy-2)^3
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so how would i do it?
@zepdrix can u help?
ok
so will it be (2x - 3y^4)3 over (4xy - 2)^3 ?
\[\large \frac{(2x-3y^4)^3(x^3+y)^0}{(4xy-2)^3}\]So you simplified it to\[\large \frac{(2x-3y^4)^3}{(4xy-2)^3}\] Yah looks good so far. hmm
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okay thanks :)
yea i got it thanks
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