Breaking a fraction down to a sum of fractions

AI Thread Summary
To rewrite the fraction 1/[(n^3)+n] as a sum of fractions, it can be expressed as (A/n) + (B/[(n^2)+1]). The initial attempt led to confusion regarding the coefficients A and B, particularly since A must equal zero while also needing to equal one. It was clarified that for the quadratic term in the denominator, the numerator should take the form Ax + B instead of just A. The correct decomposition results in the expression (1/n) - [n/(n^2 + 1)]. This highlights the importance of using the appropriate form for the numerator when dealing with irreducible quadratic factors.
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Homework Statement


Re-write the following fraction into the sum of fractions:
1/[(n^3)+n]


Homework Equations


None that I can think of. . .


The Attempt at a Solution


I first changed [(n^3)+n] to n[(n^2)+1], so by the rules, the aformentioned fraction should equate to (A/n) + (B/[(n^2)+1]). That means A * [(n^2)+1] + B * n should equate to 1. This is where I run into problems. Since there's only one n^2, that means A should equate to zero. However, there's also the constant A, which should equate to one. 0 doesn't equate to one. Is the problem faulty or am I missing something?
 
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If the quadratic term in the denominator cannot be decomposed into linear factors, then the numerator should be given the form Ax+B instead, rather than just A.
 
1/[(n^3)+n] = 1 / (n)(n^2 + 1)
= 1 + n^2 - n^2 / (n)(n^2 + 1)
= (1/n) - [n/(n^2 +1)]
 
@poster - the method u used is only valid for two linear expressions' product, but for your one you'll have to use Bn instead of just B.
 
ARGH! Totally forgot that! Thanks!
 
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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