Cubes again!

Here's a quick way to find the one number out of the choices that can be the sum of two cubes. First, we have the identity

$a^3+b^3= (a+b)( {(a+b)}^2-3ab )$

so that

$ab = \dfrac{1}{3} ( {(a+b)}^2 - n)$

where $n$ is the larger factor of any one of the 7 numbers given as choices. Those numbers can be factored into

$7 \cdot 167$
$13 \cdot 103$
$19 \cdot 67$
$5 \cdot 251$
$8 \cdot 167$
$9 \cdot 121$
$10 \cdot 113$

Only $13 \cdot 103$ yields an integer for $ab$ , so a bit more work gets us

$11^3+2^3=1339$

By the way, if you drop the requirement that the cubes be positive and throw in another cube, the problem becomes extremely difficult.

For instance, is $33$ the sum of three integer cubes? How about $42$ ?

No one knows!

$The~ numbers~are~ greater~than~ 10^3=1000, \\ So~I~ checked ~~11^3=1331~ and ~1339~ was ~greater~than ~1339.\\ So~ I~checked~1339-1331=8=2^3 !!!!!! ~~ The~ answer.\\ If~ it~ was~ not~ the~ answer~ my ~approach:\\ Let~the~two~A^3+B^3=N, A>B\\ All~\sqrt[3]{N}<11,~~so~biggest~A=10. \\ The ~cubes~are..... 1^3=1,~~~~2^3=8,~~~~3^3=27~~~~4^3=64,~~~~5^3=125,~~~~6^3=216,~~~~7^3=343,~~~~8^3=512,~~~~9^3=729.\\ None ~of~the~number~(N-1000)~ is~as~a~cub.~~So~A=10~not~possible~for~the~given~numbers. \\ Similarly~if ~for~some~N,~A=9,~(N-9^3)= (N-729)~ were ~from~~smallest=1089-729=360 ~to ~biggest=1336-729=607.~~only~cub~ within ~this~is~ 8^3=512.\\ None ~of~ the~(N-729)~gives~512.~~So~9 ~and~8~cannot~be~the~solution.~~~For~A=8,~B=8 ~N=512+512=1024~but~this~is~not~given.\\ So~there~is~no~solution~from~the~given~numbers~and~1024. ~All~numbers~are~greater~than~1024~so ~no~solution~for~other~ numbers.$