Posted by Jamie on Monday, December 17, 2012 at 1:11am.
1/(x^3)27
or
1=A(x^2+3x+9)+Bx+C(x3)
I'm trying to find the partial fraction decomposition. I think A=1/21, but how do I find the right values to plug in for x to find B and C?

Math  Steve, Monday, December 17, 2012 at 4:15am
Do a crossmultiply to put all over a common denominator:
(Ax+B)/(x^2+3x+9) + C/(x3)
= [(Ax+B)(x3) + C(x^2+3x+9)]/(x^327)
= (Ax^2+(B3A)x3B+Cx^2+3Cx+9C)/(x^327)
= [(A+C)x^2 + (B3A+3C)x + (9C3B)] / (x^327)
So, for the numerators to be identical, coefficients of like powers must be equal:
A+C = 0
B3A+3C = 0
9C3B = 1
(A,B,C) = (1/27, 6/27, 1/27)
so, 1/(x^327) = (x+6)/(27(x^2+3x+9))  1/(27(x3))
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