Finding three factors of a number with minimal sum
The problem does not require exponential time. It can be solved in subexponential, but superpolynomial time.

The first step is to find the prime factorization of $x$, which can be done with the general number field sieve.

Next, enumerate all possible values of $a,b$ and for each tuple $(a,b)$ compute $c$ to find the triple with the minumum sum. A number $x$ has at most $\exp(O(\log x / \log \log x))$ divisors, so there are also at most $\exp(O(\log x / \log \log x))$ such tuples.