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The Hessian criterion for flexes of a plane cubic applied to the defining polynomial givesConsequently the inflection points of a plane cubic on this curve are exactly its intersections with the three coordinate lines. Choose $r\in \overline{\mathbb{Q}}$ with $r^3=d$, and a primitive cube root of unity $\zeta$. The nine distinct flexes of a diagonal cubic areWith the chosen flex as identity, these points are precisely $E_d[3]$. A tangent cutting the Weil divisor $3P$ gives $3(P-O)=0$; conversely $3(P-O)=0$ makes the hyperplane Weil divisor equivalent to $3P$, which supplies a tangent with triple contact. This is the three-torsion points are flexes of a plane cubic correspondence.

On $Z=0$ only $O$ is rational, since the other cube roots of unity are not rational. On each other coordinate line a rational point exists exactly when $d$ is a rational cube, and then exactly one does. ThereforeIn particular the rational torsion subgroup cannot contain two independent points of order $3$.

One can also determine the whole rational torsion of a diagonal cubic. First it is killed by $6$. To see this, use the Weierstrass equation of an elliptic curve from (iii). At primes $\ell \equiv 2(\mathrm{mod}3)$ outside the finite set containing $2,3$ and the prime divisors of the numerator and denominator of $d$, the cube map permutes ${\mathbb{F}}_{\ell }$, sowhere the quadratic character is extended by $\chi (0)=0$, and the character sum vanishes after substituting $t=x^3-432d^2$. Reduction of torsion points on an elliptic curve injects the prime-to-$\ell$ part of rational torsion into this finite group. This injection follows because multiplication by an integer prime to $\ell$ is invertible on the formal group of an elliptic curve, and hence its reduction kernel has no such torsion.

For any prime $q>3$, choose a good $\ell \equiv 2(\mathrm{mod}3)$ with $\ell \equiv 1(\mathrm{mod}q)$; the Dirichlet theorem on primes in arithmetic progressions and the Chinese remainder theorem supply such primes outside the finite bad set. Then $q\nmid \ell +1$, excluding $q$-torsion. Similarly good $\ell \equiv 5(\mathrm{mod}12)$ exclude order $4$, and good $\ell \equiv 11(\mathrm{mod}18)$ exclude order $9$. Thus every rational torsion point has order dividing $6$.

The inversion formula in (i) shows that the only possible nonidentity rational point of order $2$ has $X=Y$ and $Z\ne 0$. It exists precisely when $(X/Z{)}^3=-d/2$, that is, when $d/2$ is a rational cube. There is at most one such point. This condition and $d$ being a cube cannot hold together, since $2$ is not a rational cube. Combining it with the three-torsion calculation gives the full answer:

Set $d=-7$. The affine rational pair $(u,v)=(2,-1)$ gives $P=(84,756)$ on the elliptic curve $y^2=x^3-21168$. It is a point of infinite order: in the rational torsion of a diagonal cubic classification, neither $-7$ nor $-7/2$ is a rational cube, so this curve has trivial rational torsion subgroup.

There is also a short direct torsion check specific to this curve. The primes $5$ and $11$ are good and have point counts $6$ and $12$, respectively, because cubing is bijective in their finite fields. Reduction at these two primes bounds the entire rational torsion by order dividing $6$, including its possible $5$- and $11$-primary parts by reduction at the other prime. The flex calculation excludes order $3$, and the inversion calculation excludes order $2$. Hence $P$ is indeed nontorsion.

Every nonzero multiple $nP$ is distinct, and its inverse projective image has $Z\ne 0$: the only rational point at infinity on the diagonal cubic is its identity. Thus each yields a distinct rational pair. More explicitly, if $nP=(x_n,y_n)$ on the Weierstrass model, thenHere $x_n\ne 0$, since $x_n=0$ would require a rational square equal to $-21168$. For example the elliptic-curve addition formula gives $2P=(28,28)$, which produces $(u_2,v_2)=(5/3,4/3)$. This proves there are infinitely many rational pairs of cubes summing to seven.