I'm not sure how to get $(1-3x^6 + 3x^{12} - x^{18}) \sum_{i=0}^{\infty} \binom{i+2}{2} x^{i}$ from $(\frac{1-x^{6}}{1-x})^3$.
I know the following series.
$$\frac{1}{1-x}=(1+x+x^2 + x^3 + x^4 + ...)$$
$$\frac{1}{1-x^6}=(1+x^6+x^{12} + x^{18} + x^{24} + ...)$$
Hint. From $$ \frac{1}{1-x}=\sum_{i=0}^{\infty}x^i,\quad |x|<1, $$ by differentiating, you get $$ \frac{1}{(1-x)^2}=\sum_{i=1}^{\infty}ix^{i-1},\quad |x|<1, $$ or, changing $i-1$ to $i$, $$ \frac{1}{(1-x)^2}=\sum_{i=0}^{\infty}(i+1)x^{i},\quad |x|<1, $$ differentiating again and changing $i-1$ to $i$ again : $$ \frac{2}{(1-x)^3}=\sum_{i=0}^{\infty}(i+1)(i+2)x^i,\quad |x|<1, $$ then dividing by $2$ observing that $$ \binom{i+2}{2} =\frac{(i+1)(i+2)}{2} $$ leads to the desired result, taking into account that $$ \left(1-x^6\right)^3=1-3 x^6+3 x^{12}-x^{18}.$$