prove $\displaystyle \sqrt{7}/^{3}\sqrt{15}$ is rational
thanks in advance.
$\displaystyle a = \frac{\sqrt{7}}{\sqrt[3]{15}} = \sqrt[6]{\frac{7^3}{15^2}} $
Assume $\displaystyle a=\frac{p}{q} $, where $\displaystyle (p,q)=1 $.
Then $\displaystyle a^6 = \frac{p^6}{q^6} $.
So $\displaystyle 7^3\cdot q^6 = 15^2\cdot p^6 $.
This means $\displaystyle 7^3\mid p^6 \implies 7\mid p $ since $\displaystyle 7 $ is prime.
So $\displaystyle 15^2\cdot p^6 =15^2\cdot (7k)^6 = 7^6\cdot 15^2\cdot k^6 = 7^3\cdot q^6\implies 7^3\cdot 15^2\cdot k^6 = q^6 $.
Therefore $\displaystyle 7 \mid q $ for the same reasons as above.
But this means $\displaystyle 7\mid (p,q) $ which is a contradiction since we were under the assumption that $\displaystyle (p,q)=1 $.
Therefore $\displaystyle \frac{\sqrt{7}}{\sqrt[3]{15}} $ is irrational.