<s>[INST] Express the real root of $x^3-3x+7$ using radicals I want to express the real root of $x^3-3x+7=0$ using radicals.

My attempt is contained in the answer below.
 [/INST] To express the real root of $x^3-3x+7$ using radicals, we can apply Cardano's method:
\begin{align*}
x^3-3x+7=0&,\text{ let } x=u+v,\quad 3uv=3\implies uv=1\\
\implies& (u+v)^3-3(u+v)+7=0\\
\implies& (u^2+2uv+v^2)(u+v)-3(u+v)+7=0\\
\implies& u^3+2u^2v+uv^2+u^2v+2uv^2+v^3-3(u+v)+7=0\\
\implies& u^3 + v^3+uv(3u+3v)-(3u+3v)+7=0\\
\implies& u^3 + v^3+(uv-1)(3u+3v)+7=0\\
\end{align*}
Since $uv=1$ we simplify to:
$$\implies u^3+v^3+7=0$$
Notice also that $uv=1\implies v=\frac1u$
$$\implies u^3+u^{-3}+7=0$$
let $w=u^3$
\begin{align*}
\implies& w+w^{-1}+7=0\\
\implies& w^2+7w+1=0\\
\implies& w=\frac{-7\pm \sqrt{49-4}}{2}\\
\implies& w=\frac{-7\pm 3\sqrt{5}}{2}\\
\end{align*}
Lets take $w=\frac{-7+ 3\sqrt{5}}{2}$, then $u=\sqrt[3]{\frac{-7+ 3\sqrt{5}}{2}}$
$$x=\sqrt[3]{\frac{-7+ 3\sqrt{5}}{2}}+ \frac{1}{\sqrt[3]{\frac{-7+ 3\sqrt{5}}{2}}}$$
$$x=\sqrt[3]{\frac{-7+ 3\sqrt{5}}{2}}+ \sqrt[3]{\frac{2}{-7+ 3\sqrt{5}}}\approx -2.426\in \Bbb R$$
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