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@hartnn @Nnesha
\[\huge ba^4 \times (2ba^4)^{-3}\]
\[\large b^42^{-3}a^{-8}\]
\(\Large (2ba^4)^{-3} = 2^{-3} b^{-3} (a^4)^{-3}\)
yes i did that
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\(b \times b^{-3} = ... ?\)
b^4
sorry b^-2
yes :) b^(-2)
(a^4)^(-3) = ... ?
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a^-12
\( a^4 \times a^{-12} \)
a^-8
\[\large \frac{ 1 }{ 8a^8b^2 }\]
yes, thats correct :)
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