A) \[1\]
B) \[2\]
C) \[3\]
D) \[4\]
Correct Answer: C
Solution :
\[\sqrt[3]{9}={{3}^{\frac{2}{3}}}\] Rationalizing factor of \[\sqrt[n]{{{a}^{p}}}\]is \[\sqrt[n]{{{a}^{n-p}}}\]or \[{{a}^{1\frac{p}{n}}}\]where \[n>p.\] Here \[\sqrt[3]{9}=\sqrt[3]{{{3}^{2}}}\]i.e., \[n=3,p=2\] R.F of \[\sqrt[3]{9}={{3}^{1\frac{2}{3}}}={{3}^{{\scriptstyle{}^{1}/{}_{3}}}}or\,\sqrt[3]{3}\] \[\therefore \] The denominator \[\sqrt[3]{9}\]when rationalised is 3.
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