Can you find the other way?

$\large\sum_{n=1}^\infty \dfrac{ \sigma_3 (n) }{n^6} = \dfrac{\zeta(a) \cdot \pi^b}c$

If the equation above holds true for positive integers $a,b$ and $c$ , find $a+b+c$ .

Notations :

• The divisor function for integer $n$ , $\sigma_k (n)$ is defined as the sum of the $k^\text{th}$ powers of the positive integers divisors of $n$ , $\displaystyle \sigma_k(n) = \sum_{d| n} d^k$ .

• $\zeta(\cdot)$ denotes the Riemann zeta function .

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I have posted one similar problem like this , but I think this problem has slightly different solutions.

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Try to solve with $d^{3}*1 = (d*1)^{2}$ .