How to find multiple variables by subtracting equations?

Question

I am working on a problem in discrete math. Specifically, I must find a closed-form solution to a recurrence relation. I found the sequence of differences and determined that the closed-form solution is a 4th degree equation. As such, I need to find the variables A, B, C, D, and E using the following equations. Unfortunately, I am not very good at basic algebra despite being in computer science.

$$ \begin{cases} 81A + 27B + 9C + 3D + E = 13 \\ 256A + 64B + 16C + 4D + E = 29 \\ 625A + 125B + 25C + 5D + E = 54 \\ 1296A + 216B + 36C + 6D + E = 90 \\ 2401A + 343B + 49C + 7D + E = 139 \\ \end{cases} $$

Answer 1

Reduce the system to four equations involving
four variables by substracting equation 1 from
equation 2, equation 2 from equation 3, equation 3
from equation 4 and equation 4 from equation 5.
What do you get?

Answer 2

1)$81A + 27B + 9C + 3D + E = 13$

2)$256A + 64B + 16C + 4D + E = 29$

3)$625A + 125B + 25C + 5D + E = 54$

4)$1296A + 216B + 36C + 6D + E = 90$

5)$2401A + 343B + 49C + 7D + E = 139$

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I: 2- 1: $175 A + 37B +7C+ D=16$

II: 3-2: $369 A+61B + 9C + D = 25$

III: 4-3: $671 A + 91B + 11C + D = 36$

IV: 5-3: $1105 A + 127 B + 13C + D = 49$.

---

X: II - I: $194A + 24B + 2C = 9$

Y: III-II: $302 A + 30B + 2C = 11$

Z: IV - III: $488 A+36B + 2C = 13$

....

$\alpha$ Y-X: $108A + 6B = 2$

$\beta$ Z-Y: $186A + 6B = 2$.

.....

$\Omega: \beta - \alpha: $ $78A = 0$

.....

$A = 0$

.....

$\alpha$ Y-X: $ 6B = 2$

$\beta$ Z-Y: $6B = 2$.

$B = \frac 13$

.....

X: II - I: $ 8 + 2C = 9$

Y: III-II: $ 10 + 2C = 11$

Z: IV - III: $12 + 2C = 13$

$C = \frac 12$

....

I: 2- 1: $ 37(\frac 13) +7(\frac 12)+ D=16$

II: 3-2: $61(\frac 13) + 9(\frac 12) + D = 25$

III: 4-3: $ 91(\frac 13) + 11(\frac 12) + D = 36$

IV: 5-3: $ 127 (\frac 13) + 13(\frac 12) + D = 49$.

$D = \frac 16$

.....

1)$ 9 + 4\frac 12 + \frac 12 + E = 13$

2)$ 21\frac 13 + 8 + \frac 23 + E = 29$

3)$ 41\frac 23 + 12\frac 12 + \frac 56 + E = 54$

4)$ 72 + 18 + 1 + E = 90$

5)$ 114\frac 13 + 24\frac 12 + 1\frac 16 + E = 139$

...

$E = -1$.