How do you write out the partial fraction decomposition of the functions:

How do you write out the partial fraction decomposition of the functions:?
All I need is the form like i have for my answers. Im not sure about a) and I know b) is definitely wrong. I would really appreciate the help. Thanks

a) x^4/(x^4-81)

Im not sure if its A/x + B/x^2 + C/X^4 + D/x^4 + E/x^4-81

b) (t^4+t^2+1)/(t^2+2)(t^2+7)^2

For this I put Ax+B/(t^2+2) + Cx+D/(t^2+7) + Ex+F/(t^2+7)^2

a)

This one is definitely wrong:

x⁴/(x⁴−81) is not A/x + B/x² + C/X⁴ + D/x⁴ + E/x⁴−81
since x, x² and x⁴ are not factors of denominator.

(x⁴−81) = (x²−9)(x²+9) = (x−3)(x+3)(x²+9)

So denominators will be (x−3), (x+3), and (x²+9)

First we need to have numerator with smaller degree than denominator:
x⁴/(x⁴−81) = (x⁴−81)/(x⁴−81) + 81/(x⁴−81)
x⁴/(x⁴−81) = 1 + 81/(x⁴−81)

So now we just need to find partial fraction decomposition of 81/(x⁴−81)
81/(x⁴−81) = A/(x−3) + B/(x+3) + (Cx+D)/(x²+9)

Multiply both sides by (x−3)(x+3)(x²+9), i.e. by (x⁴−81)
81 = A(x+3)(x²+9) + B(x−3)(x²+9) + (Cx+D)(x−3)(x+3)
81 = A(x³+3x²+9x+27) + B(x³−3x²+9x−27) + C(x³−9x) + D(x²−9)
81 = (A+B+C)x³ + (3A−3B+D)x² + (9A+9B−9C)x + (27A−27B−9D)

Matching coefficients we get:
A + B + C = 0
3A − 3B + D = 0
9A + 9B − 9C = 0
27A − 27B − 9D = 81

Solving, we get:
A = 3/4
B = −3/4
C = 0
D = −9/2

81/(x⁴−81) = (3/4)/(x−3) − (3/4)/(x+3) − (9/2)/(x²+9)

x⁴/(x⁴−81) = 1 + 81/(x⁴−81)
x⁴/(x⁴−81) = 1 + (3/4)/(x−3) − (3/4)/(x+3) − (9/2)/(x²+9)

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b)

You're almost correct for this one. Just don't mix up the variable. Use all x or all t, not a combination of both

(t⁴+t²+1)/((t²+2)(t²+7)²) = (At+B)/(t²+2) + (Ct+D)/(t²+7) + (Et+F)/(t²+7)²

Multiply both sides by (t²+2)(t²+7)²

t⁴ + t² + 1 = (At+B)(t²+7)² + (Ct+D)(t²+2)(t²+7) + (Et+F)(t²+2)
t⁴ + t² + 1 = (At+B)(t⁴+14t²+49) + (Ct+D)(t⁴+9t²+14) + (Et+F)(t²+2)

t⁴ + t² + 1 = (A+C)t⁵ + (B+D)t⁴ + (14A+9C+E)t³ + (14B+9D+F)t² + (49A+14C+2E)t + (49B+14D+2F)

Matching coefficients, we get:
A+C = 0
B+D = 1
14A+9C+E = 0
14B+9D+F = 1
49A+14C+2E = 0
49B+14D+2F = 1

Solving, we get: A = 0, B = 3/25, C = 0, D = 22/25, E = 0, F = −43/5

(t⁴+t²+1)/((t²+2)(t²+7)²) = (3/25)/(t²+2) + (22/25)/(t²+7) − (43/5)/(t²+7)²