alg 2+urgent please.
posted by Chopsticks .
2/the 6th root of 81
the answer gave an answer of "2 3rd root of 3/3"
sorry i didn't type the root signs, i dunno how.
can you show me how to work this?

the book gave me that answer.

2/the 6th root of 81
= 2/[(3^4)^(1/6)]
= 2/3^(2/3)
now multiply top and bottom by 3^(1/3)
which would give you 3^(3/3) or 3 in the bottom
= 2(3^(1/3))/3
= 2*cuberoot(3)/3, your answer.
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