A) \[\sqrt[5]{{{a}^{3}}{{b}^{2}}c}\]
B) \[\sqrt[4]{{{a}^{3}}{{b}^{2}}c}\]
C) \[\sqrt[3]{{{a}^{3}}{{b}^{2}}c}\]
D) \[\sqrt{{{a}^{3}}{{b}^{2}}c}\]
Correct Answer: A
Solution :
We have \[\sqrt[5]{{{a}^{2}}{{b}^{3}}{{c}^{4}}}\] Since, \[\sqrt[5]{{{a}^{2}}{{b}^{3}}{{c}^{4}}}\times \sqrt[5]{{{a}^{3}}{{b}^{2}}c}=\sqrt[5]{{{a}^{2}}{{a}^{3}}{{b}^{3}}{{b}^{2}}{{c}^{4}}c}\] \[\therefore \]The rationalizing factor is \[\sqrt[5]{{{a}^{3}}{{b}^{2}}c}.\]
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