A) \[\frac{10}{3}\]
B) \[-\frac{10}{3}\]
C) \[\frac{3}{15}\]
D) Both [a] and [b]
Correct Answer: D
Solution :
We have \[\sqrt[3]{3\left( \sqrt[3]{x}-\frac{1}{\sqrt[3]{x}} \right)}=2\] \[\text{Cubing both sides, we get}\] \[\text{3}\left( \sqrt[3]{x}-\frac{1}{\sqrt[3]{x}} \right)=8\Rightarrow \sqrt[3]{x}-\frac{1}{\sqrt[3]{x}}=\frac{8}{3}\] Now, squaring both sides, we get \[{{\left( \sqrt[3]{x}-\frac{1}{\sqrt[3]{x}} \right)}^{2}}=\frac{64}{9}\] Now, \[{{\left( \sqrt[3]{x}+\frac{1}{\sqrt[3]{x}} \right)}^{2}}={{\left( \sqrt[3]{x}-\frac{1}{\sqrt[3]{x}} \right)}^{2}}+4\] \[=\frac{64}{9}+4=\frac{64+36}{9}\] \[\therefore {{\left( \sqrt[3]{x}+\frac{1}{\sqrt[3]{x}} \right)}^{2}}=\frac{100}{9}\] \[\Rightarrow \left( \sqrt[3]{x}+\frac{1}{\sqrt[3]{x}} \right)=\frac{10}{3}or-\frac{10}{3}\]
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