Sorry. The problem should have been:
(x^2)/(x^3-8)(x-2)
note that x^3-8 = (x-2)(x^2+2x+4)
So, your fraction is now
x^2
--------------------
(x-2)^2 (x^2+2x+4)
As partial fractions, that becomes
A/(x-2) + B(x-2)^2 + (Cx+D)/(x^2+2x+4)
Placing all that over the common denominator, we have
A(x-2)(x^2+2x+4) + B(x^2+2x+4) + (Cx+D)(x-2)^2
= A(x^3-8)+B(x^2+2x+4) + Cx^3+(-4C+D)x^2+(4C-4D)x+4D
Expand that all out and equate coefficients with the left side (0x^3+x^2+0x+0) and we have
A+C = 0
B-4C+D = 1
2B+4C-4D = 0
-8A+4B+4D = 0
A = 1/6
B = 1/3
C = -1/6
D = 0
and you can form the fractions now.