Improving the running time of finding the count of pairs satisfying a symmetric relation
You are going in wrong direction. Entertaining the symmetry doesn't change the quadratic nature of the algorithm (using floating point in a seemingly number-theoretical problem also raises a red flag). The correct approach involves some math.

We are interested in integers \$a,b,n\$ satisfying the relation

$$\sqrt[3]{a} + \sqrt[3]{b} = \sqrt[3]{n}$$

Notice that if \$a\$ and \$b\$ have a common factor, \$n\$ necessarily has the same factor, so we are primarily interested in the solutions with \$a\$ and \$b\$ coprime.

We are going to show that such nontrivial solution is only possible if \$a\$ and \$b\$ are perfect cubes.

Now, \$(\sqrt[3]{a} + \sqrt[3]{b})^3 = n\$. Expanding the binomial gives $$3\sqrt[3]{ab}(\sqrt[3]{a} + \sqrt[3]{b}) = n - (a + b)$$

that is,

$$3\sqrt[3]{nab} = n - (a + b)$$

Since the RHS is obviously integer, so must be LHS. Assuming that \$a\$ is not a perfect cube, it must have some prime factor \$p\$ in a non-cubic power. Further assuming that \$b\$ is coprime to \$a\$, the remaining powers of \$p\$ must be supplied by \$n\$, that is \$n\$ is also divisible by \$p\$.

It means that LHS is divisible by \$p\$, and since \$b = n - a - 3\sqrt[3]{nab}\$, it follows that \$b\$ is also divisible by \$p\$, which contradicts the assumption.

We proved that the set of solutions is comprised by pairs of

$$A = x^3 z, B = y^3 z$$

with \$x, y\$ being coprime integers, and \$z\$ being an arbitrary integer. Enumerating such pairs shall not be a problem.