1. I'll give a solution just for this one: Take the reciprocal like the user above had said.

1. \(\frac{2p}{4p^2-1}\times \frac{6p+3}{6p^3} \)

2. We realize that \(\frac{6p+3}{6p^3}=\frac{2p+1}{2p^3}.\)

3. We get: \(\frac{2p\left(2p+1\right)}{\left(4p^2-1\right)\times \:2p^3}\).

4. And. \(\frac{p\left(2p+1\right)}{\left(4p^2-1\right)p^3}\), so \(\frac{2p+1}{p^2\left(4p^2-1\right)}\).

5. Factor: \(4p^2-1:\quad \left(2p+1\right)\left(2p-1\right)\)

6. Therefore, \(\frac{1}{p^2\left(2p-1\right)}\) , consequently, \(p^2\left(2p-1\right):\quad 2p^3-p^2.\)

Thus, the final answer is \(\boxed{\frac{1}{2p^3-p^2}}.\)

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