Show that $7$ divides $a^{15}-a^3$ for any integer $a$
The $5$ divisibility won't be relevant here. Just note that $a^{15}-a^3=a^{3}(a^{12}-1)$ as you did and if $7|a$ you are done, otherwise $a^6\equiv 1\mod 7$ by Fermat and so $a^{12}=(a^6)^2\equiv 1\mod 7$ so $a^{12}-1\equiv 0\mod 7$ as desired.
$a^{15}-a^3 = a^3(a^{12}-1) = a^3(a^6-1)(a^6+1)= (a^7-a)(a^9+a^3)$
$7|a^7-a$ by LFT