Posted by Tim on Friday, April 18, 2008 at 5:48pm.
Below is my problem with my answers. I don't think this is that tough, but I cannot believe that my ABC answer's are a fraction. Did I do this correctly?
Find the unknowns on the right side of the partial fraction.
(x+3)/(x^3+64)=(A)/(x+4)+(Bx+C)/(x^24x+16)
x=0
A= (x^2B4xBx+x4C+3)/(x^24x+16)
B= (x^2A+4xAx(16A+x4C+3)/(x(x+4)
C= (x^2B4xBx^2A+4x+x16A+3)/(x+4)

Algebra  Anonymous, Friday, April 18, 2008 at 6:12pm
No, I would not think you should have a fraction.

Algebra  Reiny, Friday, April 18, 2008 at 6:19pm
I multiplied both sides of your equation by (x+4)(x^2  4x + 16) which is really x^3 + 64
x+3 = A(x^2  4x + 16) + (x+4)(Bx+C)
this is an identity, so it is true for any value of x.
Let's pick "easy" values of x
let x=4
1 = A(48) + 0 > A = 1/48
let x=0
3 = 16A + 4C
3 = 16(1/48) + 4C > C = 5/6
let x=1
4 = 13A + 5B + 5C
4 = 13(1/48) + 5B + 5(5/6)
B = 1/48
Check my arithmetic, my weak point.
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