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April 17, 2015

April 17, 2015

Posted by **Liz** on Thursday, January 22, 2009 at 9:30pm.

- calculus -
**bobpursley**, Thursday, January 22, 2009 at 9:47pmI am not certian of the fractions you are using.

x^6-1 is a difference of two squares

(x^3-1)(x^3+1) Now both of those are the difference (or sum) of two cubes, which can both be factored. So you have four factors in the denominator. Frankly, I doubt if we can do much in ASCII to help you on those fractions. Notice you have a double quadratic in the denominator.

If you want a computer to assist you, put 1/(x^6-1) in the window here.

http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=algebra&s2=partial_fractions&s3=basic

- calculus -
**Liz**, Thursday, January 22, 2009 at 9:54pmI did do the sum and difference of the cubes and got these as my partial fractions:

A / (x+1)

Bx + C / (x^2-x+1)

D / (x-1)

Ex + F / (x^2+x+1)

I got the answer for A = -1/6 and D = 1/6.

But I was having trouble with the other variables...

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