Happy birthday, Ramanujan!

let P be the set of prime factors of 1729!.then: $B=\dfrac{1729!}{\phi(1729!)}=\dfrac{1729!}{1729!\prod_{p\in P} \left(1-\frac{1}{p}\right)}=\prod_{p\in P}\dfrac{p}{p-1}$ this was easily done by just the definition of the phi function.the other function given is multiplicative.proof: $f(d)=\frac{\mu^2(d)}{\phi(d)}$ we see f(1)=1, and f(ab) for coprime a,b: $f(ab)=\frac{\mu^2(ab)}{\phi(ab)}=\frac{\mu^2(a)}{\phi(a)}\frac{\mu^2(b)}{\phi(b)}=f(a)f(b)$ so it is multiplicative. the summation at 1 is 1, so it is multiplicative as well. recall that $\mu(p^k)=0$ for k>1. we can split the summation by the prime factors of 1729!(set P), or in product notation $A=\prod_{p\in P} \sum_{d|p} \left(\frac{\mu^2(d)}{\phi(d)}\right)$ why only one power of p?, because starting from 2nd power, the mobius function will return 0, so that part has no value. we can write this as $A=\prod_{p\in P} \left(\frac{\mu^2(1)}{1}+\dfrac{\mu^2(p)}{\phi(p)}\right)=\prod_{p\in P} \left(1+\dfrac{1}{p-1}\right)=\prod_{p\in P} \left(\dfrac{p}{p-1}\right)$ this is equal to B, and hence A=B.

Here is a "compressed" version of Aareyan's fine solution:

$a(n)=\sum_{d|n}\frac{\mu^2(d)}{\phi(d)}$ and $b(n)=\frac{n}{\phi(n)}$ are both multiplicative functions. For a prime power $n=p^m$ we have $a(p^m)=\frac{\mu^2(1)}{\phi(1)}+\frac{\mu^2(p)}{\phi(p)}=\frac{p}{p-1}=\frac{p^m}{p^m-p^{m-1}}=\frac{p^m}{\phi(p^m)}=b(p^m)$ Thus $a(n)=b(n)$ for all $n$ , and for $n=1729!$ we have $\boxed{A=B}$ .