Negative Fractional Exponent problem

AI Thread Summary
The discussion centers around solving the expression 3x(3x^(-1/3))^3, which is part of an old medical school entry test. The correct answer is identified as 81, with the key step being the cancellation of x terms. A participant explains that simplifying the expression leads to 27 times 3, resulting in 81, provided x is not zero. There is a mention of common mistakes, such as incorrectly placing 3^3 in the denominator. The conversation highlights the importance of careful simplification in solving fractional exponent problems.
jdoyle
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Hi All,

Can anyone walk me through this problem. This is from an old med school entry test.

The answer is 81 but I can't work out how the x terms cancel.



3x(3x^ -1/3)^3

Thanks in advance

John
 
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\frac{3x}{(^3\sqrt{x})^3}3^3=\frac{3x}{x}3^3=27*3=81

Given of course that x is not 0.
 
Hi Meldraft,

It looks so simple I don't know why I didn't do it that way. Thanks very much.

John
 
It's very easy to put that 3^3 in the denominator by accident :wink:
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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