Number of integral solutions of $\text{xyz}=3000$

$\bf{My\; Solution::}$ We Can write $x\cdot y\cdot z= 3000 = 2^3\cdot 3^1\cdot 5^3$

Now let $x=2^{x_{1}}\cdot 3^{y_{1}}\cdot 5^{z_{1}}$ and $2^{x_{2}}\cdot 3^{y_{2}}\cdot 5^{z_{2}}$ and $2^{x_{3}}\cdot 3^{y_{3}}\cdot 5^{z_{3}}$

So $x\cdot y \cdot z = 2^{x_{1}+x_{2}+x_{3}}\cdot 3^{y_{1}+y_{2}+y_{3}}\cdot 2^{z_{1}+z_{2}+z_{3}}=2^{3}\cdot 3^{1}\cdot 5^{3}$

So we get $x_{1}+x_{2}+x_{3}=3$ and $y_{1}+y_{2}+y_{3}=1$ and $z_{1}+z_{2}+z_{3} = 3\;,$

Where $0 \leq x_{1}+x_{2}+x_{3}\leq 3\;,0\leq y_{1}+y_{2}+y_{3}\leq 1\;,0\leq z_{1}+z_{2}+z_{3}\leq 3$

So we get $(x_{1}\;,x_{2}\;,x_{3}) = 10$ pairs. and $(y_{1},y_{2},y_{3}) = 3$ pairs and $(z_{1},z_{2},z_{3}) = 10$ pairs

So We Get $(x,y,z) = 10 \times 3 \times 10 = 300$ positive integer ordered pairs.

Now It is possible that any two variables is $(-)$ve and one is $(+)$ve.

So We get Total ordered pairs is $ = \bf{all\; positive}+\bf{any\; two \; is \; negative.}$

Note that $3000 = 2^3 \cdot 3^1 \cdot 5^3$. Let $x=2^{x_1}3^{x_2}5^{x_3}$, $y=2^{y_1}3^{y_2}5^{y_3}$ and $z=2^{z_1}3^{z_2}5^{z_3}$. Then obtain the number of solutions to $$x_1+x_2+x_3 = 3$$ $$y_1+y_2+y_3 = 1$$ $$z_1+z_2+z_3 = 3$$ The product of the number of non-negative solutions will give us the total number of solutions. Hence, in our case, the number of non-negative solutions to the first one is $\dbinom{3+3-1}{3-1} =10$, the second one is $\dbinom{1+3-1}{3-1} =3$ and the third one is $\dbinom{3+3-1}{3-1} =10$. Hence, total number of solutions is $10\cdot 3 \cdot 10 = 300$.