Right triangles $T_1$ and $T_2$ have areas $2$ and $3$, respectively. A side of $T_1$ is congruent to a side of $T_2$, and a different side of $T_1$ is congruent to a different side of $T_2$. What is the square of the product of the lengths of the other (third) sides of $T_1$ and $T_2$?

1. If $T_2$ has legs $a,b$ and $T_1$ has leg $a$ and hypotenuse $b$: $ab=2A_2$ and $a\sqrt{b^2-a^2}=2A_1$.
2. Setting $u=a^2,v=b^2$: $uv=4A_2^2$ and $u(v-u)=4A_1^2$, giving $u^2=4(A_2^2-A_1^2)$.
3. With $A_1=2,A_2=3$: $u^2=4(9-4)=20$, $v=\frac{4\cdot9}u=\frac{36}u$. The desired quantity is $v^2-u^2=\dfrac{4A_2^4}{A_2^2-A_1^2}-4(A_2^2-A_1^2)=\dfrac{324}5-20=\boxed{\dfrac{224}5}$.