How do you express (8x-1)/(x^3 -1)8x−1x3−1 in partial fractions?
1 Answer
Dec 8, 2016
Explanation:
Note that
So:
(8x-1)/(x^3-1) = A/(x-1) + (Bx+C)/(x^2+x+1)8x−1x3−1=Ax−1+Bx+Cx2+x+1
color(white)((8x-1)/(x^3-1)) = (A(x^2+x+1) + (Bx+C)(x-1))/(x^3-1)8x−1x3−1=A(x2+x+1)+(Bx+C)(x−1)x3−1
color(white)((8x-1)/(x^3-1)) = ((A+B)x^2+(A-B+C)x+(A-C))/(x^3-1)8x−1x3−1=(A+B)x2+(A−B+C)x+(A−C)x3−1
Equating coefficients we find:
{ (A+B = 0), (A-B+C = 8), (A-C = -1) :}
Adding all three equations, we find:
3A = 7
So:
A=7/3
From the first equation we can deduce:
B = -A = -7/3
From the third equation:
C = A+1 = 7/3+1 = 10/3
So:
(8x-1)/(x^3-1) = 7/(3(x-1)) - (7x-10)/(3(x^2+x+1))
