, 2013

Using the coefficients $a_0, a_1, a_2, \dots, a_9$, we can write
$$(x+1)^3(x+2)^3(x+3)^3 = a_0 + a_1x + a_2x^2 + \dots + a_8x^8 + a_9x^9.$$
Substitution of $x = 1$ and $x = -1$ in the equation above yields, respectively
$$a_0 + a_1 + a_2 + \dots + a_8 + a_9 = 2^3 \cdot 3^3 \cdot 4^3$$
$$a_0 - a_1 + a_2 - a_3 + a_4 - a_5 + a_6 - a_7 + a_8 - a_9 = 0.$$
From these identities we obtain $a_0 + a_2 + a_4 + a_6 + a_8 = \frac{1}{2}(2^3 \cdot 3^3 \cdot 4^3 + 0) = 6912$. By substituting $x = 0$ in the equation above, we also get $a_0 = 1^3 \cdot 2^3 \cdot 3^3 = 216$. We can therefore conclude that $a_2 + a_4 + a_6 + a_8 = 6912 - 216 = 6696$.