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Find the value of the product of $\sqrt 3 $ and $\sqrt[3]{2}$.

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Last updated date: 25th Apr 2024
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Answer
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Hint: write the roots in terms of exponents.
We are supposed to find the product of $\sqrt 3 $ and $\sqrt[3]{2}$. So, the first step is to express roots in terms of exponents.
Therefore,
$\sqrt 3 \times \sqrt[3]{2} = {3^{\dfrac{1}{2}}} \times {2^{\dfrac{1}{3}}}$
Now, the step is to make the powers same,
Therefore,
$\sqrt 3 \times \sqrt[3]{2} = {3^{\dfrac{1}{2} \times \dfrac{3}{3}}} \times {2^{\dfrac{1}{3} \times \dfrac{2}{2}}}$
$\sqrt 3 \times \sqrt[3]{2} = {3^{\dfrac{3}{6}}} \times {2^{\dfrac{2}{6}}}$
We can take $\left( {{3^3}} \right)$ and $\left( {{2^2}} \right)$ in one bracket,
Therefore,
$\sqrt 3 \times \sqrt[3]{2} = {\left( {{3^3}} \right)^{\dfrac{1}{6}}} \times {\left( {{2^2}} \right)^{\dfrac{1}{6}}}$
We can write $\left( {{3^3}} \right) = 27$ and$\left( {{2^2}} \right) = 4$,
Therefore,
$\sqrt 3 \times \sqrt[3]{2} = {27^{\dfrac{1}{6}}} \times {4^{\dfrac{1}{6}}}$
We can take${\left( {27 \times 4} \right)^{\dfrac{1}{6}}}$ by the formula ${a^m} \times {b^m} = {(ab)^m}$
$\sqrt 3 \times \sqrt[3]{2} = {108^{\dfrac{1}{6}}}$
Now we can write it as,
$\sqrt 3 \times \sqrt[3]{2} = \sqrt[6]{{108}}$
Answer $ \Rightarrow \sqrt[6]{{108}}$
Note: make sure in the end while writing the answer, you express it in terms of root.