Riemann Zeta Function of 6/Proof 2

Proof
Dividing out the x factor on both sides and equating the product with the sum, we have:

Equating the $x^2$ term on both sides of the equation, we obtain the value of the sum of the individual terms in the Basel Problem:

Equating the $x^4$ term on both sides of the equation, we obtain the value of the sum of the product of every unique combination of $2$ terms from the Basel Problem:

Equating the $x^6$ term on both sides of the equation, we obtain the value of the sum of the product of every unique combination of $3$ terms from the Basel Problem:

When we take the cube of a sum, we have:

Let $A = \dfrac 1 {1^2}, B = \dfrac 1 {2^2}, C = \dfrac 1 {3^2}, \cdots $

Then the (Cube of Sum) becomes:

and the first term on the (Sum of Cubes) becomes:

To obtain the remaining two terms on the, we have:

Finally, we have: