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Solve x^3 = 64/27.
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there is a rule: \(\LARGE\color{black}{ \left(\begin{matrix} \frac{a}{b} \\ \end{matrix}\right)^c=\frac{a^c}{b^c} }\)
I was thinking: ±8/3 or 8/3
you know that \(\LARGE\color{black}{ 27=3^3 }\) and that \(\LARGE\color{black}{ 64=4^3 }\)
Oh, 4/3
yes.
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well, can it not be ±4/3?
nvm
yes, it is not \(\LARGE\color{black}{ \pm }\)
k
it would be \(\large\color{black}{ \pm }\) when dealing with even powers.
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Submitting ~crosses fingers~
90%
Nice;)
yep.
Yw
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