Can an integer that is $3\pmod 7$ be expressed as a sum of two cubes?

Hint: Use Fermat little theorem:

If $7\nmid x$ then $$ x^6\equiv 1 \pmod 7 $$ and so $$ x^3\equiv \pm 1 \pmod 7 $$

Note that $x^3\equiv0,\pm1\bmod7$ for any $x\in\mathbb Z$. So, $x^3+y^3=0,\pm1,\pm2\bmod7$, and hence, this is not possible.

The answer is no. For every integer $z$, $z^3 \mod 7=0,1,-1$. The sum of any two of these is not $3 \mod 7$.