Prove that $5^{1/3}+7^{1/2}$ is irrational 
Goal: Prove that $5^{1/3}+7^{1/2}$ is irrational.

Idea: We can prove this is irrational by supposing it is rational and finding a contradiction. So, $5^{1/3}+7^{1/2} = p/q$ where $p$ and $q$ are integers that have no factors in common other than $1$. The issue here is that I can not seem to find a way manipulate this equation to a form where I can find contradiction. I tried taking each side to the power of 6... That was an unmanageable mess.  Any ideas?

Edit:  Swapped the cube root and the square root in the display.  Fixed.  This also altered the subsequent argument.
To use your method, you never want to raise to a power anything that has both $\sqrt[3]{5}$ and $\sqrt{7}$ in it.  So, you isolate these one at a time.
Suppose $\sqrt[3]{5} + \sqrt{7} \in \Bbb{Q}$ so there are $p,q \in \Bbb{Z}$ with $q \neq 0$ such that $\sqrt[3]{5} + \sqrt{7} = p/q$.  By reducing $p/q$ to lowest terms, we may assume that $\gcd(p,q) = 1$.  Since solving quadratics is harder than solving linears (for the second isolation), we start by cubing.  We compute $$\begin{align}
    \sqrt[3]{5} &= p/q - \sqrt{7},  \\
    5 &= (p/q - \sqrt{7})^3 \\
      &= p^3/q^3 - p^2\sqrt{7}/q^2 + 7p/q - 7\sqrt{7}, \text{ and}\\
    \sqrt{7} &= \frac{5-p^3/q^3 - 7p/q}{-p^2/q^2 - 7} \\
    7 &= \left( \frac{5q^3 - p^3 - 7pq^2}{-p^2q - 7q^3} \right)^2, \text{ so} \\
    7q^2 \left( p^2 + 7q^2 \right)^2 &= \left( p^3 + 7pq^2 - 5q^3 \right)^2 .
\end{align}$$
Now work modulo $4$.  The squares are congruent to $0$ or $1$, so the left-hand side is either $0$ or $3 \pmod{4}$ and the right-hand side is either $0$ or $1 \pmod{4}$.  So, either $q$ or $p^2 + 7q^2$ is even and $p^3 + 7pq^2 - 5q^3$ is even.  
If $q$ is even, $p^3 + 7pq^2 - 5q^3 \cong p^3 \pmod{4}$.  Since $p^3 + 7pq^2 - 5q^3$ is even, we must have $p$ is even, contradicting our assumption that $p/q$ was reduced to lowest terms.  
Otherwise, we have $p^2 + 7q^2$ is even.  Note that $p^3 + 7pq^2 - 5q^3 = p(p^2+7q^2)-5q^3$, so is even only if $q$ is even.  But then $p^2 + 7q^2$ is even only if $p$ is and again we contradict reduction to lowest terms.
