if $a^3 + 2b^3\equiv 0 \pmod {27} $, then $a\equiv b \equiv 0 \pmod 3$ if  $a^3 + 2b^3\equiv 0 \pmod {27}$, then $a\equiv b \equiv 0 \pmod 3$
could you please show this. I am just able show $a\equiv b \pmod 3$

HInt $\rm\ 3\nmid b\,\Rightarrow\, mod\ 9\!:\ b^{-1}\ exists\,\Rightarrow\,\color{#c00}{(a/b)^{3}\equiv -2},\,$ contradicting $-2$ is not a cube mod $9$.
Similarly for the $\rm\color{#c00}{reciprocal}$ case $\,\rm3\nmid a,\,$ because $\,\color{#c00}{-1/2}\equiv 8/2\equiv 4$ is not a cube. 
