# Show that \sqrt[3]{\pi} is transcendental

Show that $\sqrt[3]{\pi }$ is transcendental

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ritmesysv

Step 1
Suppose $a=\sqrt[n]{\pi }$ is the root of a polynomial $P\left(x\right)$ with rational coefficients of positive degree.
Writing $P\left(a\right)=0$ and using that ${a}^{n}=\pi$, powers ${a}^{k}$ with $k>n$ can be replaced with ${a}^{k}={\pi }^{\frac{k}{n}}·{\sqrt[n]{\pi }}^{k\mathrm{mod}n}$
It follows that there exist rational polynomials in  such that:
1) ${a}_{n-1}{\sqrt[n]{\pi }}^{ n-1}+{a}_{n-2}{\sqrt[n]{\pi }}^{ n-2}+\cdots +{a}_{1}\sqrt[n]{\pi }+{a}_{0}=0$
Multiply the above by $\sqrt[n]{\pi }$ successively $n-1$ times:
2) ${a}_{n-2}{\sqrt[n]{\pi }}^{ n-1}+{a}_{n-3}{\sqrt[n]{\pi }}^{ n-2}+\cdots +{a}_{0}\sqrt[n]{\pi }+{a}_{n-1}\pi =0$
${a}_{n-3}{\sqrt[n]{\pi }}^{ n-1}+{a}_{n-2}{\sqrt[n]{\pi }}^{ n-2}+\cdots +{a}_{n-1}\pi \sqrt[n]{\pi }+{a}_{n-2}\pi =0\cdots$
${a}_{0}{\sqrt[n]{\pi }}^{ n-1}+{a}_{n-1}\pi {\sqrt[n]{\pi }}^{ n-2}+\cdots +{a}_{2}\pi \sqrt[n]{\pi }+{a}_{1}\pi =0$
Considering the n equations (1) and (2) as a linear system in  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 $\sqrt{n}\left\{\pi \right\}$ 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 $P\left(\sqrt[n]{\pi }\right)=0⇒R\left(\pi \right)=0$ where R is the resultant $R\left(z\right)=\text{res}\left(P\left(x\right),,{x}^{n}-z,,x\right)$, only without assuming prior knowledge of resultants or other higher algebra results.)