Is $\frac{\sqrt7}{\sqrt[3]{15}}$ rational or irrational? Is $\frac{\sqrt7}{\sqrt[3]{15}}$ rational or irrational? Prove it.
I am having a hard time with this question. So far what I did was say, assume it's rational, then $$\frac{\sqrt7}{\sqrt[3]{15}}=\frac{x}{y} \Rightarrow \sqrt{7}y=x\sqrt[3]{15}$$
I then showed the product of a rational number and an irrational number is irrational so the expression above is irrational on the left and right side. I can't get to a contradiction to prove it's irrational so I'm currently thinking it is rational but I don't know how to go about proving it.
 A: Suppose the number is rational, let
$$ \frac{\sqrt{7}}{\sqrt[3]{15}} = \frac{x}{y} $$
where $\gcd(x, y) = 1$, then 
$$ \sqrt{7}y = \sqrt[3]{15} x \implies 7^3 y^6 = 225 x^6 $$
So 
$$ 7^3  | x^6     $$
which implies $ 7| x $.
Let $x = 7m, \; m \in \mathbb N$. Then we have 
$$  7^3 y^6 = 225 \times 7^6 m^6 $$
which means 
$$  y^6 = 225 \times 7^3 m^6 $$
once again we have 
$$ 7^3  | y^6     $$
which implies $ 7| y $. And this is contradictory to the coprimality between $x$ and $y$.
A: If $x$ is rational then so is $x^3$, and hence so is $\frac{15}{7}x^3$.
Applying this to
$$x=\frac{\sqrt7}{\root3\of{15}}$$
shows that if $x$ is rational then so is $\sqrt7$.  But I expect you already know that this is not the case.  Therefore $x$ is irrational.
(If you are not allowed to assume that $\sqrt7$ is irrational, it is pretty easy to prove.)
A: Factoring $\,7^{\,\large \color{}{3}} y^{\large 6}\!=15^{\large 2}x^{\large\color{}{6}}$ into primes, $\,7\,$ has odd power $\,3\!+\!6j\,$ on LHS, but even $\,6k\,$ on RHS 
A: Hint: This is the sixth root of $\dfrac{343}{225}$.
A: hint:$7^3y^6 = 15^2x^6 \Rightarrow \left(\dfrac{x}{y}\right)^6 = r \Rightarrow t^6 - r = 0 \Rightarrow pt^6 - q = 0$. Use Eisenstein's criteria.
