How to Verify the 5th Term of the Binomial Expansion of (3-2/x)^9?

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To verify the 5th term of the binomial expansion of (3-2/x)^9, the general term formula is applied: T(r+1) = n!/(n-r)!r! * A^(n-r) * B^r. For this case, r is set to 4, A is 3, B is -2/x, and n is 9. The calculated term results in -489888/x^4. To check the answer, one should ensure the correct application of the formula and confirm the sign of each component in the expansion.
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hi there I am abit stuck here. i got a q saying :
in binomial expansion of (3-2/x)^9 find the 5th term using the general term of the binomial expansion and check your answer

(3-2/x)^9

used formula

=N!/(n-r)!r! * A^(n-r) * b ^ r

r= 4
a= 3
b= - 2/x
n=9

got a answer of -489888/x^4

How do i go about checking my answer??
 
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My answer is like yours, but without the negative sign.
 
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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