If $$x=\sqrt[3]{a+\sqrt{a^{2}+b^{3}}}$$ + $$\sqrt[3]{a-\sqrt{a^{2}+b^{3}}}$$, then $$x^{3}+3bx$$ is equal to

Expression : $$x=\sqrt[3]{a+\sqrt{a^{2}+b^{3}}}$$ + $$\sqrt[3]{a-\sqrt{a^{2}+b^{3}}}$$

Cubing both sides, we get:

=> $$x^3 = (\sqrt[3]{a + \sqrt{a^2 + b^3}})^3 + \sqrt[3]{a - \sqrt{a^2 + b^3}})^3 + 3(\sqrt[3]{a+\sqrt{a^{2}+b^{3}}}) (\sqrt[3]{a-\sqrt{a^{2}+b^{3}}})[\sqrt[3]{a+\sqrt{a^{2}+b^{3}}} + \sqrt[3]{a-\sqrt{a^{2}+b^{3}}}]$$

=> $$x^3 = a + \sqrt{a^2 + b^3} + a - \sqrt{a^2 + b^3} + 3(a^2 - a^2 - b^2)^\frac{1}{3}[x]$$

=> $$x^3 = 2a + (-3bx)$$

$$\therefore$$ $$x^3 + 3bx = 2a$$