Question

Express the rational denominator:
\[\dfrac{{\sqrt 2 .\sqrt[3]{3}}}{{\sqrt[3]{3} + \sqrt 2 }}\]

Answer 1

Hint: The rationalization is the multiplying the denominator term with opposite sign to the numerator and the denominator simultaneously. For suppose if the denominator is given as (x+y) and asked to rationalize the denominator, then we need to multiply (x-y) to the numerator and the denominator.

Complete step-by-step answer:
 In the given question we have to rationalize the denominator \[\sqrt[3]{3} + \sqrt 2 \], it means we have to multiply the term \[\sqrt[3]{3} - \sqrt 2 \] to the numerator and the denominator of the problem.
The given expression is \[\dfrac{{\sqrt 2 .\sqrt[3]{3}}}{{\sqrt[3]{3} + \sqrt 2 }}\].
 If we rationalize the denominator \[\sqrt[3]{3} + \sqrt 2 \], we get,
\[\begin{array}{c}
\dfrac{{\sqrt 2 .\sqrt[3]{3}}}{{\sqrt[3]{3} + \sqrt 2 }} = \dfrac{{\sqrt 2 .\sqrt[3]{3}}}{{\sqrt[3]{3} + \sqrt 2 }} \times \dfrac{{\sqrt[3]{3} - \sqrt 2 }}{{\sqrt[3]{3} - \sqrt 2 }}\\
 = \dfrac{{\left( {\sqrt 2 .\sqrt[3]{3}} \right)\sqrt[3]{3} - \left( {\sqrt 2 .\sqrt[3]{3}} \right)\sqrt 2 }}{{\left( {\sqrt[3]{3} + \sqrt 2 } \right)\left( {\sqrt[3]{3} + \sqrt 2 } \right)}}\\
 = \dfrac{{\sqrt 2 {{.3}^{\dfrac{2}{3}}} - 2\sqrt[3]{3}}}{{{3^{\dfrac{2}{3}}} - 2}}
\end{array}\]
Therefore, the rationalization of \[\dfrac{{\sqrt 2 .\sqrt[3]{3}}}{{\sqrt[3]{3} + \sqrt 2 }}\] is \[\dfrac{{\sqrt 2 {{.3}^{\dfrac{2}{3}}} - 2\sqrt[3]{3}}}{{{3^{\dfrac{2}{3}}} - 2}}\].

Note: Here, the terms are little complex because the terms consist of roots with power of more than 2, so you should be sure about the rationalizing and do not make mistakes while multiplying the complex root terms. The numerator and the denominators are the positions of the given terms in the division, do not get confuse between the both, the numerator is the term that placed at the top of the division (here \[\sqrt 2 .\sqrt[3]{3}\] is the numerator term) and the denominator is the term that placed at the bottom of the division (it means \[\sqrt[3]{3} + \sqrt 2 \] is the denominator term). While rationalizing the denominator be aware that the terms in the denominator is related with addition symbol then we have changed the symbol as subtraction to rationalize and the terms in the denominator.