Question

Find the product of $\sqrt[3]{35}$ and $\sqrt{2}$.

Answer 1

The product of the given surds can be determined as follows:

$\sqrt[3]{35}\times \sqrt{2}=5^{\frac{1}{3}}\times 2^{\frac{1}{2}}$

Now, $LCM$ of $3$ and $2$ is $6$, therefore,

$\sqrt[3]{35}\times \sqrt{2}=5^{\frac{1}{3}}\times 2^{\frac{1}{2}}=5^{\frac{1}{3}\times \frac{6}{6}}\times 2^{\frac{1}{2}\times \frac{6}{6}}=5^{\frac{2}{6}}\times 2^{\frac{3}{6}}=\sqrt[6]{65^2}\times \sqrt[6]{62^3}=\sqrt[6]{65^2\times 2^3}=\sqrt[6]{625\times 8}=\sqrt[6]{6200}$.

Hence, $\sqrt[3]{35}\times \sqrt{2}=\sqrt[6]{6200}$.