Selmer's cubic is locally soluble but globally insoluble

Consider Selmer's cubic $3X^3+4Y^3+5Z^3=0$. It has a real point because the one-variable equation $3x^3+5=0$ has a real root, giving $(x,0,1)$. It has a $2$-adic point because $f(X)=3X^3+9$ satisfies $f(1)\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}2)$ and $f^{\mathrm{\prime }}(1)=9\equiv ̸0(\mathrm{m}\mathrm{o}\mathrm{d}2)$, so [L1] lifts the mod-$2$ root and yields a point $(x,1,1)$ in ${\mathbb{Z}}_2^3$. It has a $3$-adic point because $80\equiv -1(\mathrm{m}\mathrm{o}\mathrm{d}27)$, so for $g(Y)=Y^3-80$ one has $\mathrm{\mid }g(-1){\mathrm{\mid }}_3=3^{-4}<3^{-2}=\mathrm{\mid }{g}^{\mathrm{\prime }}(-1){\mathrm{\mid }}_3^2$; [L2] therefore gives a $3$-adic root $y$ of $g$, and then $(0,y,-4)$ lies on the cubic. It has a $5$-adic point because $h(X)=3X^3+32$ satisfies $h(1)\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}5)$ and $h^{\mathrm{\prime }}(1)=9\equiv ̸0(\mathrm{m}\mathrm{o}\mathrm{d}5)$, so [L1] yields a $5$-adic root $x$, giving the point $(x,2,0)$.

Conrad's cited note proves that Selmer's cubic has local points over every remaining ${\mathbb{Q}}_p$ but no nontrivial rational point over $\mathbb{Q}$. Thus the curve is locally soluble at every completion while globally insoluble, refuting any naive extension of Hasse-Minkowski from quadratic forms to cubic curves.