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Let $A$, $B$, and $C$ be constants so that \[\frac{1}{x^3-x^2-21x+45}=\frac{A}{x+5}+\frac{B}{x-3} + \frac{C}{(x - 3)^2}\]holds for all real numbers $x$ other than $-5$ and $3.$ What is $A$?

 Jun 6, 2024
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x^3 - x^2 -21x + 45  factors as   (x-3)^2 (x + 5)

Multiply through by this factorization  and  we  get

 

1  =  A(x-3)^2 + B (x-3) ( x + 5)  + C(x + 5)

 

Let x = -5

Then  A(-8)^2  = 1  →  A = 1/64

 

cool cool cool

 Jun 6, 2024
edited by CPhill  Jun 7, 2024

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