Question

-1/(x^3)-27

or

-1=A(x^2+3x+9)+Bx+C(x-3)

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?

Answer 1

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))