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> Find the total number of integer solutions $(x,y,z)\in\mathbb{Z}$ of the equation $xyz=24$.
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I have tried $xyz = 2^3 \cdot 3$

**My Process:**

Factor $x$, $y$, and $z$ as
$$
\begin{cases}
x = 2^{x\_1} \cdot 3^{y\_1}\\
y = 2^{x\_2} \cdot 3^{y\_2}\\
z = 2^{x\_3} \cdot 3^{y\_3}
\end{cases}
$$

Now $2^{x\_1+x\_2+x\_3} \cdot 3^{y\_1+y\_2+y\_3} = 2^3 \cdot 3$, so

$$
\begin{cases}
x\_1+x\_2+x\_3 = 3\\
y\_1+y\_2+y\_3 = 1
\end{cases}
$$

where $x\_1,x\_2,x\_3,y\_1,y\_2,y\_3\geq 0$. Thus the number of solutions for the first is $\binom{5}{2}=10$ and second is $\binom{3}{2}=3$. This gives a total number of positive solutions of $10 \cdot 3 = 30$

How can I calculate the total number of solutions? Is this process correct?