From Förberedande kurs i matematik 1

In power form, the expressions become

\displaystyle \begin{align} \sqrt{2}\bigl(\sqrt[4]{3}\bigr)^{3} &= 2^{1/2}\bigl(3^{1/4}\bigr)^{3} = 2^{1/2}3^{3/4},\\[5pt] \sqrt[3]{2}\cdot 3 &= 2^{1/3}3^{1}\,\textrm{.} \end{align}

Admittedly, it is true that \displaystyle 2^{1/2} > 2^{1/3} and \displaystyle 3^1 > 3^{3/4}, but this does not help us to say anything about how the products are related to each other. Instead, we observe that the exponents 1/2, 3/4, 1/3 and 1 have \displaystyle 3\cdot 4 = 12 as the lowest common denominator which we can take out

\displaystyle \begin{align} 2^{1/2}3^{3/4} &= 2^{6/12}3^{(3\cdot 3)/12} = \bigl(2^{6}\cdot 3^{9}\bigr)^{1/12},\\[5pt] 2^{1/3}3^{1} &= 2^{4/12}3^{12/12} = \bigl(2^{4}\cdot 3^{12}\bigr)^{1/12}\,\textrm{.} \end{align}

Now, we can compare the bases \displaystyle 2^6\cdot 3^9 and \displaystyle 2^4\cdot 3^{12} with each other and so decide which number is larger.

Because

\displaystyle \frac{2^6\cdot 3^9}{2^4\cdot 3^{12}} = 2^{6-4}3^{9-12} = 2^{2}3^{-3} = \frac{2^{2}}{3^{3}} = \frac{4}{27} < 1

the denominator \displaystyle 2^{4}\cdot 3^{12} is larger than the numerator \displaystyle 2^6\cdot 3^9, which means that \displaystyle \sqrt[3]{2}\cdot 3 is larger than \displaystyle \sqrt{2}\bigl(\sqrt[4]{3}\bigr)^{3}.