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evaluate (a^4)^3 when a = (2r3)^1r6 Ill rewrite it neater

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\[\huge(a ^{4})^{3} \]
\[ \huge a= (\frac{ 2 }{ 3 })^{\frac{ 1 }{ 6 }}\]

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Other answers:

i just sub a equal whatever in a^12 but i can never get it right
(a^4)^3 = a^(4*3) (a^4)^3 = a^(12) So if you want to find (a^4)^3, then you can find a^(12) instead since they are the same
\[ \Large a= (\frac{ 2 }{ 3 })^{\frac{ 1 }{ 6 }}\] \[ \Large a^{12}\] \[ \Large \left((\frac{ 2 }{ 3 })^{\frac{ 1 }{ 6 }}\right)^{12}\] \[ \Large (\frac{ 2 }{ 3 })^{\frac{ 1 }{ 6 } * 12}\] \[ \Large (\frac{ 2 }{ 3 })^{\frac{ 12 }{ 6 }}\] \[ \Large (\frac{ 2 }{ 3 })^{2}\] I'll let you finish
so the final answer is 4r9?
wow thanks so much anyways
you mean 4/9 right?
not sure how/where the r comes into play
btw \[\Large \frac{4}{9} = 4/9\]
oh lol in my calc to write something over something its something r something :P
oh interesting, never seen or heard of that before
actaully some question written in australia are written 8r9 which means 8/9
i gotcha, that makes more sense now
never ever seen that notation before

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