If p=\sqrt{(a+\sqrt{a^{2}+b^{3}})}+\sqrt{(a-\sqrt{a^{2}+b^{3}})}, then what is p^{3}+3bp equal to?

Explanation

Using (x+y)^3 = x^3+y^3+3xy(x+y). Let x, y be the cube roots. x^3+y^3 = 2a. xy = \sqrt{a^2 - (a^2+b^3)} = -b. Thus, p^3 = 2a - 3bp, which gives p^3 + 3bp = 2a.

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