If the sum of reciprocals of cubes of all the triangular numbers converges to the number \(A\),

\[A = 1 + \dfrac{1}{3^3} + \dfrac{1}{6^3} + \dfrac{1}{{10}^3} + \dfrac{1}{{15}^3} + \dfrac{1}{{21}^3} + \dfrac{1}{{28}^3} + \dfrac{1}{{36}^3} + \dfrac{1}{{45}^3} + \cdots \]

and the sum of reciprocals of squares of all the triangular numbers converges to the number \(B\),

\[B = 1 + \dfrac{1}{3^2} + \dfrac{1}{6^2} + \dfrac{1}{{10}^2} + \dfrac{1}{{15}^2} + \dfrac{1}{{21}^2} + \dfrac{1}{{28}^2} + \dfrac{1}{{36}^2} + \dfrac{1}{{45}^2} + \cdots \]

then \(\dfrac AB\) can be expressed as \(M \dfrac{\left( P - \pi^2 \right)}{\left( \pi^2 - O \right)}\) where \(M, O\) and \(P\) are positive integers.

Find \(M+O+P\).

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