Question

The product of \[\sqrt[3]{2}\]and \[\sqrt 2 \] is
A. \[\sqrt[6]{{32}}\]
B. \[\sqrt[6]{{24}}\]
C. \[\sqrt[6]{{16}}\]
D. \[\sqrt[6]{4}\]

Answer 1

Hint: First we will rewrite the roots of the number \[\sqrt[3]{2}\] and \[\sqrt 2 \] in the fraction form and when the bases in the above expression are same, then the base is taken common and powers are added. Then we will rewrite the fractions in the power into the root form to find the required value.

Complete step by step answer:

We are given that \[\sqrt[3]{2} \times \sqrt 2 \].

Rewriting the roots of the number \[\sqrt[3]{2}\] and \[\sqrt 2 \] in the fraction form, we get

\[ \Rightarrow {2^{\dfrac{1}{3}}} \times {2^{\dfrac{1}{2}}}\]

We know when the bases in the above expression are same, then the base is taken common and powers are added, that is, \[{a^n} \cdot {a^m} = {a^{m + n}}\], we get
\[
   \Rightarrow {2^{\dfrac{1}{3} + \dfrac{1}{2}}} \\
   \Rightarrow {2^{\dfrac{{2 + 3}}{6}}} \\
   \Rightarrow {2^{\dfrac{5}{6}}} \\
 \]

Using the property of powers that \[{\left( {{a^m}} \right)^n} = {a^{mn}}\] in the above equation, we get
\[
   \Rightarrow {\left( {{2^5}} \right)^{\dfrac{1}{6}}} \\
   \Rightarrow {32^{\dfrac{1}{6}}} \\
 \]

Rewriting the fractions in the power into the root form, we get
\[ \Rightarrow \sqrt[6]{{32}}\]

Thus, the required value is \[\sqrt[6]{{32}}\]
Hence, option A is the right answer

Note: In solving this question, students must know that when the bases are same then their values are added, but when the bases are different, then you cannot multiply the terms. Some students try to solve the product in the root form, which gets even harder. Do not multiply the powers of the same bases while multiplying, the answer will be wrong.