Posted by Jamie on Monday, December 17, 2012 at 1:11am.
Do a cross-multiply to put all over a common denominator:
(Ax+B)/(x^2+3x+9) + C/(x-3)
= [(Ax+B)(x-3) + C(x^2+3x+9)]/(x^3-27)
= (Ax^2+(B-3A)x-3B+Cx^2+3Cx+9C)/(x^3-27)
= [(A+C)x^2 + (B-3A+3C)x + (9C-3B)] / (x^3-27)
So, for the numerators to be identical, coefficients of like powers must be equal:
A+C = 0
B-3A+3C = 0
9C-3B = -1
(A,B,C) = (1/27, 6/27, -1/27)
so, -1/(x^3-27) = (x+6)/(27(x^2+3x+9)) - 1/(27(x-3))
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