It's well-known that quadratic forms over the rational numbers $\mathbf{Q}$ satisfy the Hasse-Minkowski theorem. This is to say that they are isotropic over $\mathbf{Q}$ if and only if they are isotropic over $\mathbf{R}$ and over $\mathbf{Q}_p$ for all prime numbers $p$.

Of course the same statement does not hold for cubic forms, as demonstrated by Selmer, who showed that the ternary cubic form $3x^3 + 4y^3 + 5z^3$ has nontrivial zeros over $\mathbf{R}$ and over $\mathbf{Q}_p$ for all $p$. We say then that Selmer's example *violates the Hasse Principle*.

Now I saw a claim recently, which seems a little too good to be true, that if you take a *ternary* cubic form which violates the Hasse Principle then you can show that it must be diagonal.

Now I don't want to talk about approaches to this or potential proofs or anything of the sort. Merely, when I saw this I thought there ought to be some easy counterexample, but I couldn't immediately come up with one. If you're like me, when you think of Hasse Principle counterexamples, you first think of Selmer's example or of a hyperelliptic example, or maybe something that's just harder to write down. No obvious non-diagonal cubic forms lie in that bunch. As such, I pose the following question.

Is there a non-diagonal ternary cubic form which violates the Hasse Principle? Can you write it down explicitly?