Bump in the Right Direction: Simplifying b^2-a^2

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The expression b^2 - a^2 simplifies to (b - a)(b + a), which is known as the difference of squares. Dividing by (b - a) is valid only when a is not equal to b. Participants discuss the importance of understanding polynomial division in this context. Clarification is provided that the simplification process is valid under specific conditions. Overall, the discussion emphasizes the mathematical principles behind the difference of squares and polynomial division.
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b^2-a^2
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b-a simplifies to a+b
I didn't think I could simply divide because of the subtraction.
Is that how this is simplified ?
If not then please give me a bump in the right direction.
 
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expand this for me ...

(a^2-b^2)=?

"Difference of squares ..."
 
Find out what you get by multiplying b-a and b+a.
 
do you know polynomial division? (btw, your result is valid only when a is not equal to b)
 
Argggh! Thanks everyone!
 
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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