We have three equations, Eq1, Eq2, Eq3.

Eq1 rsa = r * s * a

Eq2 r_s_a = r + s + a

Eq3 rs_ra_sa = r * s + r * a + s * a

1.Factor Eq3 for r

rs_ra_sa = r * (s + a) + s * a

2. Rewrite Eq1 + Substitute:

rsa = r * s * a

s * a = rsa/r

rs_ra_sa = r * (s + a) + rsa/r

3. Rewrite Eq2 + substitute

s + a = r_s_a -r

rs_ra_sa = r * (r_s_a -r) + rsa/r

4. Rewrite

rs_ra_sa = r * (r_s_a -r) + rsa/r

rs_ra_sa = -r^2 + r * r_s_a + rsa/r

-r^2 + r * r_s_a + rsa/r - rs_ra_sa = 0

-r^3 + r^2 * r_s_a - rs_ra_sa *r + rsa = 0

4. Now we can solve the cubic equation using sympy to find r, s, a. They are the three roots of this equation as all previous rewriting were symmetric and could have been done also with "s" and "a".

from sympy import var, Eq, solve

```python
from sympy import var, Eq, solve
a = -1
b = r_s_a
c = -rs_ra_sa
d = rsa
x = var('x')
sol = solve(Eq(a*x**3 + b*x**2 + c*x + d, 0), x)
print(sol)
r, s, a = sol
```

Now we can compute d and solve the challenge:

```python
from sympy import mod_inverse

phi = (r-1) * (s-1) * (a-1)
d = mod_inverse(e, phi)

flag_enc = pow(flag_enc, d, rsa)
print(int.to_bytes(flag_enc, 54))
```
Solution:
srdnlen{W3lc0m3_t0_srdnlen_ctf_2023_crypt0_ch4ll3ng3s}