Answer: Understanding Exponents: Inside vs Outside Brackets

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The discussion clarifies the application of exponents in expressions involving negative bases and parentheses. It highlights that the placement of parentheses is crucial to avoid ambiguity, as seen in the examples provided. Specifically, the expression (-y^2)^2 simplifies to (-y)^4, while (-y^2)^3 leads to a different interpretation without proper parentheses. The correct notation, such as ((-y)^2)^2, ensures clarity and accurate results. Ultimately, understanding how to properly use parentheses is essential for correctly evaluating expressions with exponents.
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This is probably a very easy question but it is messing me up.

(-y^2)^2 = (-y)^4

but

(-y^2)^3 = (-y^6)

why does one exponent need to be inside the brackets and the other outside?

Example: (-5^2)^2 = 625 = (-5)^4

(-5^2)^3 = -15625 = (-5^6)
 
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You really need to move your parentheses to avoid amibiguity; your first line should really read -(y2)2=-(y4)
 
cristo said:
You really need to move your parentheses to avoid amibiguity; your first line should really read -(y2)2=-(y4)

but that would make the answer -625, not 625
 
Sorry, I read it wrong. Well, in that case, you need more parentheses, since -x^2 is very ambiguous; I would take it to mean -(x^2). You should write ((-y)^2)^2, which is equal to (-y)^4, using the correct exponent rule.
 
Ok, so ((-y)^2)^3 = (-y)^6
 
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Yes. Notice that it is also equal to y^6 if you square the inside bracket first, since (-y)^2=y^2
 
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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