If $N=2^5\cdot 3^4 \cdot 5^3$, then number of ways in which $N$ be split into $2$ factors such that their L.C.M. is $\dfrac{N}{6}$

Question

If $N=2^5\cdot 3^4 \cdot 5^3$, then number of ways in which $N$ be split into $2$ factors such that their L.C.M. is $\dfrac{N}{6}$

My Approach: $\dfrac{N}{6}=2^{4}\cdot {3^3}\cdot5^3$

Let two numbers $a=2^{\alpha_{1}}\cdot3^{\beta_{1}}\cdot5^{\gamma_{1}}$ and $b=2^{\alpha_{2}}\cdot3^{\beta_{2}}\cdot5^{\gamma_{2}}$

Also $\alpha_{1}+\alpha_{2}=5$, $\beta_{1}+\beta_{2}=4$, $\gamma_{1}+\gamma_{2}=3$

And at-least one of $\alpha_{1}, \alpha_{2}$ must be $4$, at least one of $\beta_{1},\beta_{2}$ must be 4, and at least one of $\gamma_{1},\gamma_{2}$ must be $3$

Now I don't know how to proceed further?

Better method to solve these kind of problems are welcomed