partial fraction

x³ + 9x² +27x +27
= (x)³ + 3(x)²(3) + 3(x)(3)² + (3)³
= (x + 3)³

Hence, (x² + 4x + 8)/(x³ + 9x² +27x +27) = (x² + 4x + 8) / (x + 3)³

Let (x² + 4x + 8)/(x³ + 9x² +27x +27) = [A/(x + 3)] + [B/(x + 3)²] + [C/(x + 3)³]

Then, x² + 4x + 8 = A(x + 3)² + B(x + 3) + C

Put x = -3 :
(-3)² + 4(-3) + 8 = C
C = 5
Hence, x² + 4x + 8 = A(x + 3)² + B(x + 3) + 5

Put x = 0 :
(0)² + 4(0) + 8 = A(0 + 3)² + B(0 + 3) + 5
9A + 3B = 3 ...... [1]

Put x = -2 :
(-2)² + 4(-2) + 8 = A(-2 + 3)² + B(-2 + 3) + 5
A + B = -1
3A + 3B = -3 ...... [2]

[1] - [2] :
6A = 6
A = 1

Put A = 1 into [1] :
9(1) + 3B = 3
3B = -6
B = -2

Hence, (x² + 4x + 8)/(x³ + 9x² +27x +27)
= [1/(x + 3)] - [2/(x + 3)²] + [5/(x + 3)³]

x^2 + 4x + 8 除以 x + 3 得 x + 1 餘 5,
x + 1 除以 x + 3 得 1 餘 -2.
所以 x^2 + 4x + 8 = (x + 3)^2 - 2(x + 3) + 5
即
(x^2 + 4x + 8) / (x^3 + 9x^2 + 27x + 27)
= [(x + 3)^2 - 2(x + 3) + 5] / (x + 3)^3
= 1/(x + 3) - 2/(x + 3)^2 + 5/(x + 3)^3