Precalculus, Chapter 7, 7.4, Section 7.4, Problem 56

x^3/((x+2)^2(x-2)^2)
Let x^3/((x+2)^2(x-2)^2)=A/(x+2)+B/(x+2)^2+C/(x-2)+D/(x-2)^2
x^3/((x+2)^2(x-2)^2)=(A(x+2)(x-2)^2+B(x-2)^2+C(x-2)(x+2)^2+D(x+2)^2)/((x+2)^2(x-2)^2)
x^3/((x+2)^2(x-2)^2)=(A(x+2)(x^2-4x+4)+B(x^2-4x+4)+C(x-2)(x^2+4x+4)+D(x^2+4x+4))/((x+2)^2(x-2)^2)
x^3/((x+2)^2(x-2)^2)=(A(x^3-4x^2+4x+2x^2-8x+8)+B(x^2-4x+4)+C(x^3+4x^2+4x-2x^2-8x-8)+D(x^2+4x+4))/((x+2)^2(x-2)^2)
x^3/((x+2)^2(x-2)^2)=(A(x^3-2x^2-4x+8)+B(x^2-4x+4)+C(x^3+2x^2-4x-8)+D(x^2+4x+4))/((x+2)^2(x-2)^2)
x^3/((x+2)^2(x-2)^2)=(x^3(A+C)+x^2(-2A+B+2C+D)+x(-4A-4B-4C+4D)+8A+4B-8C+4D)/((x+2)^2(x-2)^2)
:.x^3=x^3(A+C)+x^2(-2A+B+2C+D)+x(-4A-4B-4C+4D)+8A+4B-8C+4D
equating the coefficients of the like terms,
A+C=1
-2A+B+2C+D=0
-4A-4B-4C+4D=0
8A+4B-8C+4D=0
Now we have to solve the above four equations to get the solutions of A,B,C and D.
Rewrite the third equation as,
-4(A+C)-4B+4D=0
substitute the expression of ( A+C) from the first equation in the above equation,
-4(1)-4B+4D=0
-4B+4D=4
4(-B+D)=4
-B+D=1
D=1+B
Express C in terms of A from first equation,
C=1-A
Substitute the expressions of C and D in the second equation,
-2A+B+2(1-A)+1+B=0
-2A+B+2-2A+1+B=0
-4A+2B=-3 (equation 5)
Substitute the expressions of C and D in the fourth equation,
8A+4B-8(1-A)+4(1+B)=0
8A+4B-8+8A+4+4B=0
16A+8B=4
4(4A+2B)=4
4A+2B=1 (equation 6)
Solve equation 5 and 6 to get the solutions of A and B,
Add equation 5 and 6,
4B=-3+1
4B=-2
B=-2/4
B=-1/2
Plug the value of B in the equation 5,
-4A+2(-1/2)=-3
-4A-1=-3
-4A=-3+1
-4A=-2
A=1/2
Now plug the values of A and B in the expressions of C and D,
C=1-1/2
C=1/2
D=1+(-1/2)
D=1/2
:.x^3/((x+2)^2(x-2)^2)=1/(2(x+2))-1/(2(x+2)^2)+1/(2(x-2))+1/(2(x-2)^2)