Simplifying Powers and Parentheses: (27-3y6)1/3 Solution

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The discussion focuses on simplifying the expression (27-3y6)1/3. Participants agree that the expression can be rewritten as (1/273y6)1/3, emphasizing the importance of evaluating (1/273). They reference exponent rules, particularly how to distribute exponents across terms. The final result simplifies to y^2/27, confirming the correct approach to the problem. Overall, the conversation highlights key exponent rules for simplifying complex expressions.
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Homework Statement


(27-3y6)1/3


The Attempt at a Solution


I THINK THIS IS HOW YOU SHOULD DO THIS:
(1/273Y6)1/3

My feeling is the (1/273) should still be evaluated. Yes?
 
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Remember that ## (a^{x})^{y} = a^{xy} ## and ## a^{-x}=\frac{1}{a^x} ##

How does this help you?
 
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Distribute the exponent

$$(x^a \! y^b)^c=x^{a c} \! y^{b c} \\
(5^8 \! s^4)^{(1/4)}=5^{8 (1/4)} \! s^{4 (1/4)}=5^2 \! s$$
 
solution

ok.
I get y2/27
 
datafiend said:
ok.
I get y2/27

You got it.
 
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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