Show that $\sqrt[3]{\pi}$ is transcendental A similar question about $\sqrt{\pi}$ was asked here and was beautifully explained. Now, how do we show that $\sqrt[3]{\pi}$ is also transcendental using similar techniques, without any advanced machinery in field theory? More generally, how to prove that $\sqrt[n]{\pi}$ is transcendental?
Any hint would be greatly appreciated-many thanks in advance!
 A: Suppose $\,a=\sqrt[n]{\pi}\,$ is the root of a polynomial $\,P(x)\,$ with rational coefficients of positive degree.
Writing $\,P(a)=0\,$ and using that $\,a^n = \pi\,$, powers $\,a^k\,$ with $\,k \gt n\,$ can be replaced with $\,a^k = \pi^{\lfloor k / n\rfloor} \cdot \sqrt[n]{\pi}^{\,k \bmod n}\,$. It follows that there exist rational polynomials in $\,\pi\,$ $\,a_j \,\big|\,_{j=0,1,\dots,n-1}\,$ such that:
$$
a_{n-1} \sqrt[n]{\pi}^{\,n-1} + a_{n-2} \sqrt[n]{\pi}^{\,n-2}+\dots + a_1 \sqrt[n]{\pi} + a_0 = 0 \tag{1}
$$
Multiply the above by $\,\sqrt[n]{\pi}\,$ successively $\,n-1\,$ times:
$$
a_{n-2} \sqrt[n]{\pi}^{\,n-1}+a_{n-3} \sqrt[n]{\pi}^{\,n-2}+\dots + a_0 \sqrt[n]{\pi} + a_{n-1}\pi = 0
\\ a_{n-3} \sqrt[n]{\pi}^{\,n-1}+a_{n-2} \sqrt[n]{\pi}^{\,n-2}+\dots + a_{n-1}\pi\, \sqrt[n]{\pi} + a_{n-2}\pi = 0
\\ \dots
\\ a_{0} \sqrt[n]{\pi}^{\,n-1}+a_{n-1}\pi\, \sqrt[n]{\pi}^{\,n-2}+\dots + a_2 \pi\, \sqrt[n]{\pi} + a_1\pi = 0 \tag{2}
$$
Considering the $\,n\,$ equations $\,(1)\, \& \,(2)\,$ as a linear system in $\,1, \sqrt[n]{\pi}, \sqrt[n]{\pi}^{\,2}, \dots, \sqrt[n]{\pi}^{\,n-1}\,$, it is a homogeneous system with a non-trivial solution, so its determinant must be zero. But all coefficients are rational polynomials in $\,\pi\,$ which implies $\,\pi\,$ is algebraic. The contradiction means that the original assumption cannot hold true, so $\,\sqrt[n]{\pi}\,$ is transcendental.
The same line of proof works for the $\,n^{th}\,$ root of any transcendental number, not just $\,\pi\,$.
(The above is essentially proving that $\,P\left(\sqrt[n]{\pi}\right)=0 \implies R(\pi) = 0\,$ where $\,R\,$ is the resultant $R(z) = \text{res}\big(P(x), \,x^n - z, \,x\big)$, only without assuming prior knowledge of resultants or other higher algebra results.)
