Prove that $a^3+b^3=10^5$ has no positive integer solution. 
Prove that there do not exist two positive integers, $a$ and $b$, such that $$a^3+b^3=100\,000$$

I tried to use congruence modulo $7$ and some other modulo, but it does not seem to work. If possible, please prove it with modular congruences.

The possible cubic residues modulo $19$ are:
$$1,7,8,11,12,18$$
Further, $100000\equiv 3\pmod{19}$
The only two residues whose sum is equivalent to $3$ would be $11$ and $11$.
This implies that both $a$ and $b$ are one of $5\pmod{19},16\pmod{19}$ and $17\pmod{19}$
Further suppose that $a\geq b$.  This implies that $a^3\geq 50000$ and $b^3\leq 50000$
Since $(16+2\cdot 19)^3>100000$ and since $(17+19)^3<50000$, this would imply that $a^3=(5+2\cdot 19)^3=79507$ as this is the only cube of a number from one of those equivalence classes in the desired range but then $b^3=100000-79507=20493$ is not a perfect cube.  Hence, no solutions exist.
