Infinite Sum from Radicals

Note that $\text{rad}(n)$ basically means that the numbers only has $2$ , $3$ , and $5$ as their prime factors. Thus the number is of the form $2^a\cdot 3^b\cdot 5^c$ .

We want to find the value of $\sum_{a=1}^{\infty} \sum_{b=1}^{\infty}\sum_{c=1}^{\infty}\dfrac{1}{(2^a\cdot 3^b\cdot 5^c)^2}=\sum_{a=1}^{\infty} \sum_{b=1}^{\infty}\sum_{c=1}^{\infty}\dfrac{1}{2^{2a}\cdot 3^{2b}\cdot 5^{2c}}$

However, $\sum_{a=1}^{\infty} \sum_{b=1}^{\infty}\sum_{c=1}^{\infty}\dfrac{1}{2^{2a}\cdot 3^{2b}\cdot 5^{2c}}=\left(\sum_{a=1}^{\infty}\dfrac{1}{2^{2a}}\right)\left(\sum_{b=1}^{\infty}\dfrac{1}{3^{2b}}\right)\left(\sum_{c=1}^{\infty}\dfrac{1}{5^{2c}}\right)=\left(\dfrac{\frac{1}{4}}{1-\frac{1}{4}}\right)\left(\dfrac{\frac{1}{9}}{1-\frac{1}{9}}\right)\left(\dfrac{\frac{1}{25}}{1-\frac{1}{25}}\right)=\dfrac{1}{576}$

Thus our answer is $\boxed{576}$ .