Discussion

\[ \begin{aligned} x &= \left(a+\sqrt{a^{2}+b^{3}}\right)^{1/3} \\ &+ \left(a-\sqrt{a^{2}+b^{3}}\right)^{1/3} \end{aligned} \]

then the value of \(x^{3}+3bx-2a\) is :

Answer: C

The value is 0.

Write \(x = p + q\), where \(p\) and \(q\) are the two cube roots. Then:

\[ \begin{aligned} p^{3} + q^{3} &= 2a \\ pq &= \left(a^{2} - (a^{2} + b^{3})\right)^{1/3} \\ &= (-b^{3})^{1/3} = -b \end{aligned} \]

Cubing \(x = p + q\):

\[ \begin{aligned} x^{3} &= p^{3} + q^{3} + 3pq(p + q) \\ &= 2a - 3bx \end{aligned} \]

Hence \(x^{3} + 3bx - 2a = 0\). The other options, \(a^{2}\), \(a^{2} + b^{3}\) and 1, would leave terms that the identity cancels completely.

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