Expanding, we have \[a^6-2a^5+9a^4+2a^3+6a^2+9>0 \iff a^4(a-1)^2+8a^4+2a^3+6a^2+9>0.\]Note that $8a^4+2a^3\ge 0$ as the function $f(x)=8x^4+2x^3$ has $f'(x)=32x^3+6x^2$ which equals to zero when $x=0, -\frac{3}{16}$, checking gives $f(0)=0$ as the minimum value. So as $a^4(a-1)^2\ge 0, 8a^4+2a^3\ge 0, 6a^2\ge 0, 9>0$, we have the inequality holds for all real number $a$. $\quad \blacksquare$