$$3240=2^3\cdot3^4\cdot5^1$$
The factor $2$ can be split among $3$ divisors in $10$ different ways:
- $0,0,3$
- $0,1,2$
- $0,2,1$
- $0,3,0$
- $1,0,2$
- $1,1,1$
- $1,2,0$
- $2,0,1$
- $2,1,0$
- $3,0,0$
The factor $3$ can be split among $3$ divisors in $15$ different ways:
- $0,0,4$
- $0,1,3$
- $0,2,2$
- $0,3,1$
- $0,4,0$
- $1,0,3$
- $1,1,2$
- $1,2,1$
- $1,3,0$
- $2,0,2$
- $2,1,1$
- $2,2,0$
- $3,0,1$
- $3,1,0$
- $4,0,0$
The factor $5$ can be split among $3$ divisors in $3$ different ways:
Hence the total number of ways to write it as a product of $3$ divisors is $10\cdot15\cdot3=450$.
You are almost right. As already observed in the comments, note that, calling $k_1, k_2, k_3... $ the exponents of each prime number in the factorization, if your final product $(k_1+1)(k_2+1)(k_3+1)... $ is odd, then this means that all exponents are even and so the number is a perfect square. In this case, to find the total nunber of ways to express the number as a product of two factors (regardless of whether these two factors are equal or different) you would have to add $ 1$ to your final product and divide to 2, since the symmetric distributions of the prime numbers in two factors $a \cdot b $ include one where $a=b $. For example, for $36= 2^2 \cdot 3^2$, we have $3 \cdot 3=9$ "total" divisors ($1, 2, 3, 4, 6, 9, 12, 18, 36$) but $5$ ways to express $36$ as a product of two factors ($36 \cdot 1, 18 \cdot 2, 12 \cdot 3, 9 \cdot 4, 6 \cdot 6$).
So if the question talks about two "different" factors, you have to subtract 1 and divide to 2 (in this example we have to exclude the product $6 \cdot 6$).
In contrast, if the final product is even, then the number is not a perfect square and so you only have to divide to 2.
Best Answer
We're trying to count factorizations $126=abc$ with $a\le b\le c$. Note that, except for $1\cdot1\cdot126$ and $3\cdot3\cdot14$, the rest have $a\lt b\lt c$, since $3$ is the only repeated prime factor of $126$. Let $m$ denote the number with $a\lt b\lt c$, so the number we want to find is $m+2$.
If we remove the condition $a\le b\le c$, the number of factorizations $126=abc$ is $6m+3\cdot2=6m+6$. But it's also $3\cdot6\cdot3=54$, since the $2$ and $7$ can be assigned to any of the three factors and the pair of $3$'s can be assigned in $6$ different ways. Setting $6m+6=54$, we find $m=8$, so the answer we're after is $8+2=10$.