$\left ( 1+x^{2}+x^{4}+x^{6}+x^{8}+.... \right )\left ( 1+x^{4}+x^{8}+x^{12}+.... \right )\left ( 1+x^{6}+x^{12}+x^{18}+.... \right )$
= $(\dfrac{1}{(1-x^2)})^1 .(\dfrac{1}{(1-x^4)})^1. (\dfrac{1}{(1-x^6)})^1$
=$\sum_{k_1 = 0}^{\infty }\binom{n+k_1-1}{k_1}(x^2)^{k_1}.\sum_{k_2 = 0}^{\infty }\binom{n+k_2-1}{k_2}(x^4)^{k_2}.\sum_{k_3 = 0}^{\infty }\binom{n+k_3-1}{k_3}(x^6)^{k_3}$
$\because \ (\dfrac{1}{1-x})^n \ = \ \sum_{k = 0}^{\infty }\binom{n+k-1}{k}(x)^{k}$
= $\sum_{k_1 = 0}^{\infty }\binom{k_1}{k_1}(x^2)^{k_1}.\sum_{k_2 = 0}^{\infty }\binom{k_2}{k_2}(x^4)^{k_2}.\sum_{k_3 = 0}^{\infty }\binom{k_3}{k_3}(x^6)^{k_3}$
= $\sum_{k_1 = 0}^{\infty }(x^2)^{k_1}.\sum_{k_2 = 0}^{\infty }(x^4)^{k_2}.\sum_{k_3 = 0}^{\infty }(x^6)^{k_3}$
$k_1 = 1 , k_2 = 2, k_3 = 0$
$= (x^2)(x^8). = 1.x^{10}$ , we get the $co-efficient \ of \ x^{10} \ is \ 1$
$k_1 = 0 , k_2 = 1, k_3 = 1$ , we get the $co-efficient \ of \ x^{10} \ is \ 1$
$k_1 = 2 , k_2 = 0, k_3 = 1$ , we get the $co-efficient \ of \ x^{10} \ is \ 1$
$k_1 = 5 , k_2 = 0, k_3 = 0$ , we get the $co-efficient \ of \ x^{10} \ is \ 1$
$k_1 = 3 , k_2 = 1, k_3 = 0$ , we get the $co-efficient \ of \ x^{10} \ is \ 1$
On adding we get $1 + 1 + 1 + 1 + 1 = 5$