# `x^3/((x + 2)^2(x - 2)^2)` Write the partial fraction decomposition of the rational expression. `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,

`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)`

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