(If you spot a mistake in my solution please let me know in the comments below.)
No, there is no initial configuration nor set of moves that allows us to reach the goal. Note how \(x \equiv x^3 \mod 2\). That is, \(x\) has the same parity as \(x^3\). Let us assume that, at a given point, all the integers in the bag add up to \(S\). We show that the parity of the sum of all integers in the bag doesn't change when we remove the cards \(a,b\) and then add the card \(a^3+b^3\):
\[ x \equiv x^3 \mod 2 \implies S \equiv S - a - b + a^3 + b^3 \iff S + a + b \equiv S + a^3 + b^3 \mod 2\]
Thus we can't end up only with \(73\) in the bag, as everything in the bag should add up to an even number and \(73\) is odd.
Bonus question: find a solution that would still work for \(2, 74, 308\) (and for an infinity of even numbers), even though those are even.
Let me know what you think in the comment section below!