<s>[INST] Decomposing $\frac{1 - 8z + 27z^2 - 35z^3 + 14z^4}{(1-z)^2(1-2z)^2(1-3z)}$ I got stuck at this expression and couldn't decompose it further. The book I am referring to has the answer $$\frac{1 - 8z + 27z^2 - 35z^3 + 14z^4}{(1-z)^2(1-2z)^2(1-3z)} = \frac5{4(1-z)} + \frac1{2(1-z)^2} - \frac3{1-2z} - \frac2{(1-2z)^2} + \frac{17}{4(1-3z)}$$ But I am not able to derive the same. Can anyone please help me out on how to resolve it?
 [/INST] Was able to decompose it using following method
$let \frac{1−8z+27z^2−35z^3+14z^4}{(1−z)^2(1−2z)^2(1−3z)}=\frac{A}{(1-z)}+\frac{B}{(1-z)^2}+\frac{C}{(1-2z)}+\frac{D}{(1-2z)^2}+\frac{E}{(1-3z)}
$
Multiply both sides by $(1−z)^2(1−2z)^2(1−3z)$
$1−8z+27z^2−35z^3+14z^4 = A(1-z)(1-2z)^2(1-3z)+
\\B(1-2z)^2(1-3z)+
\\C(1-z)^2(1-2z)(1-3z)+
\\D(1-z)^2(1-3z)+
\\E(1-z)^2(1-2z)^2\\$
$= A(1-8z+23z^2-28z^3+12z^4) + \\B(1-7z+16z^2-12z^3)+\\C(1-7z+17z^2-17z^3+6z^4)+\\D(1-5z+7z^2-3z^3)+\\E(1-6z+13z^2-2z^3+4z^4)\\$
$ =(A+B+C+D+E)-\\(8A+7A+7C+5D+6E)z+\\(23A+16B+17C+7D+13E)z^2-\\(28A+12B+17C+3D+12E)z^3+\\(12A+6C+4E)z^4\\$
Comparing LHS with RHS gives us 5 equations:
$A+B+C+D+E=1$
$8A+7B+7C+5D+6E=8$
$23A+16B+17C+7D+13E=27$
$28A+12B+17C+3D+12E=35$
$12A+6C+4E=14$
Solving these 5 equation in 5 variables gives { A = 5/4, B = 1/2, C = -3, D = -2, E = 17/4 }
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