We have m = sqrt(3) − sqrt(2) and n = sqrt(3) + sqrt(2).

We have $m = \sqrt{3} - \sqrt{2}$ and $n = \sqrt{3} + \sqrt{2}$.
Using the binomial expansion $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$:

$$m^3 = (\sqrt{3})^3 - 3(\sqrt{3})^2(\sqrt{2}) + 3(\sqrt{3})(\sqrt{2})^2 - (\sqrt{2})^3$$ $$m^3 = 3\sqrt{3} - 3(3)\sqrt{2} + 3\sqrt{3}(2) - 2\sqrt{2}$$ $$m^3 = 3\sqrt{3} - 9\sqrt{2} + 6\sqrt{3} - 2\sqrt{2}$$ $$m^3 = 9\sqrt{3} - 11\sqrt{2}$$

Using the binomial expansion $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:

$$n^3 = (\sqrt{3})^3 + 3(\sqrt{3})^2(\sqrt{2}) + 3(\sqrt{3})(\sqrt{2})^2 + (\sqrt{2})^3$$ $$n^3 = 3\sqrt{3} + 3(3)\sqrt{2} + 3\sqrt{3}(2) + 2\sqrt{2}$$ $$n^3 = 3\sqrt{3} + 9\sqrt{2} + 6\sqrt{3} + 2\sqrt{2}$$ $$n^3 = 9\sqrt{3} + 11\sqrt{2}$$

From Step 3 of the previous response, we established that $m \Delta n = n^3 - m^3$.

$$m \Delta n = (9\sqrt{3} + 11\sqrt{2}) - (9\sqrt{3} - 11\sqrt{2})$$ $$m \Delta n = 9\sqrt{3} + 11\sqrt{2} - 9\sqrt{3} + 11\sqrt{2}$$ $$m \Delta n = (9\sqrt{3} - 9\sqrt{3}) + (11\sqrt{2} + 11\sqrt{2})$$ $$m \Delta n = 0 + 22\sqrt{2}$$ $$m \Delta n = \boxed{22\sqrt{2}}$$