A) \[2{{a}^{3}}\]
B) \[-\,\,2{{a}^{3}}\]
C) \[1\]
D) \[0\]
Correct Answer: D
Solution :
Given, \[x={{(a+\sqrt{{{a}^{2}}+{{b}^{3}}})}^{1/3}}+{{(a-\sqrt{{{a}^{2}}+{{b}^{3}}})}^{\frac{1}{3}}}\] On cubing both sides, we get \[{{x}^{3}}=(a+\sqrt{{{a}^{2}}+{{b}^{3}}})+(a-\sqrt{{{a}^{2}}+{{b}^{3}}})\] \[+\,\,3\,\,{{(a+\sqrt{{{a}^{2}}+{{b}^{3}}})}^{1/3}}{{(a-\sqrt{{{a}^{2}}+{{b}^{3}}})}^{1/3}}\] \[\{{{(a+\sqrt{{{a}^{2}}+{{b}^{3}}})}^{1/3}}+{{(a-\sqrt{{{a}^{2}}+{{b}^{3}}})}^{1/3}}\}\] \[\Rightarrow \] \[{{x}^{3}}=2a-3b\,\,(x)\] \[\Rightarrow \]\[{{x}^{3}}+3bx-2a=0\]
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