Pretty trivial to use modern graphing software, so I assume a purely pencil and paper method is desired.
Assume a<-3. Rewrite the left side into a^6 + [(-a-3)*a^2] + [6*(-a-3)] + 1.
-a-3 is positive for all a<-3 so then the rewritten expression of the left side is the sum of four positive terms. Therefore there are no "large" negative solutions.
Assume a>6. Rewrite the left side into [(a^2-4)*a^4] + [(a-2)*a^3] + [(a^2-6)*a^2] + [(a-6)a^3] + [a^4-17].
It is easy to check that for a>3 then each difference a^2-4, a^2-6, a-6, and a^4-17 are always positive; so then the rewritten expression of the left side is the sum of four positive terms. Therefore there are no "large" positive solutions.
Then the possible integer values of a which satisfy the inequality are limited to the interval of -3 to 5.
a=-3 -> f(a)=730
a=-2 -> f(a)=51
a=-1 -> f(a)=-14
a=0 -> f(a)=-17
a=1 -> f(a)=-30
a=2 -> f(a)=-5
a=3 -> f(a)=586
a=4 -> f(a)=3831
a=5 -> f(a)=15178
From this list the solution set is {-1, 0, 1, 2}.
Solution