I believe the solution is
U =1, V=7, C= 4, W=2, X=5
Inelegantly, we know the sum of the digits in UVCUVWX is a cube
The maximum value this can have is with U or V equal to 9, and the other duplicate equal to 8, then the other digits being 7, 6, and 5. This is a sum of 52. The minimum it can be is 0+0+1+1+2+3+4 which is 11. The only cube in this range is 27.
Candidates for WX are constrained to be two digit squares and candidates for UV are constrained to be two digit primes.
We want at least one digit to be a 7 or an 8 to get enough digits for the answer. Since they are all distinct, there can be no repeats. Combinations with primes in the 80s don't allow the target of 27 for the sums of the digits because of the repeats; we need 2U+2V+C+W+X=27 and with UV=83 we need C+W+X =3 and no two digit square fits the requirements for WX. A similar issue occurs for 89, so we want something smaller. Two digit primes with 7 are 17, 37, 67, 71, and 73 (97 omitted because 9^7 plus other terms will be too large). 17 or 71 will combine with 25 as candidates for UV and WX respectively and force C to be 4, and checking the exponentiation, we need UV =17.
Edited to fix typo