= Solution
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 $Z\ne0$: 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>, then
$$
\boxed{u_n=\frac{y_n+252}{6x_n},\qquad v_n=\frac{252-y_n}{6x_n},\qquad u_n^3+v_n^3=7\quad(n\ge1).}
$$
Here $x_n\ne0$, 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 \b[there are infinitely many rational pairs of cubes summing to seven].
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