Prove $(3+\sqrt{11})^{1/3}$ is irrational. I can't say I've gotten very far. You can show $3 + \sqrt{11}$ is irrational, call it $a$. Then I tried supposing it's rational, i.e.:
$a^{1/3}$ = $\frac{m}{n}$ for $m$ and $n$ integers.
You can  write $m$ and $n$ in their canonical factorizations, then cube both sides of the equation...but I can't seem to derive a contradiction. 
 A: If $(3 + \sqrt{11})^\frac{1}{3}=\frac{p}{q}$ were a rational, then $3 + \sqrt{11}=\frac{p^3}{q^3}$ would also be a rational. 
A: If you denote $x = (3 + \sqrt{11})^{\frac 13}$, you can show that the equation $x^6 - 6x^3 - 2 = 0$ holds and appeal to Rational Root Theorem.
Rational Root Theorem
A: Rational Root Theorem

If $P(x) = a_nx^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \cdots + a_1x + a_0$, where all of the $a_i$ are integers, and $P\left( \dfrac uv \right)=0$ where $u$ and $v$ are integers, then $v \mid a_n$ and $u \mid a_0$.

Let $x = (3 + \sqrt{11})^{\frac 13}$. Then
\begin{align}
   x^3 &= 3 + \sqrt{11} \\
   x^3-3 &= \sqrt{11}\\
   x^6 - 6x^3 + 9 &= 11 \\
   x^6 - 6x^3 - 2 &= 0 \\
\end{align}
According to the rational root theorem, the only possible rational zeros of the polynomial $P(x) = x^6 - 6x^3 - 2$ are $1, -1, 2,$ and $-2$.
We find $P(1) = -7$, $P(-1)=-5$, $P(2)=14$ and $P(-2)=110$
Hence $P(x) = x^6 - 6x^3 - 2$ has no rational roots. It follows that 
$x = (3 + \sqrt{11})^{\frac 13}$ is irrational.
