Qyestion Find Coefficient of $x^{12}$ in $${(x*\frac{\left(1-x^6\right)}{(1-x)})}^3$$
My attempt (using Generator Functions)
Since we know that, $$\frac{\left(1-x^6\right)}{(1-x)} = \left(x^5+x^4+x^3+x^2+x+1\right)$$
therefore the given problem reduces to finding the co-efficient of $x^{12}$ in
$(x*\left(x^5+x^4+x^3+x^2+x+1\right))^3$,
taking that $x^{3}$ out, that means we need to find the co-efficient of $x^{9}$ in
$\left(x^5+x^4+x^3+x^2+x+1\right)^3$,
which in turn is equivalent to $\sum_{m=0}^{\infty}\binom{m+2}{2}x^{m}$, the coefficient of $x^{9}$ will be $\binom{11}{2}. $ i.e $55$
(i haven't shifted the series to match the correct order though but it isn't necessary right as that is a variable change i.e substituing $m+2 = r$)
but the (book's) answer says the co-efficient isn't that, rather it's 25.
Where did i go wrong in applying the logic?
Thanks a lot for your time.
(I did check other amazing answers here, they all seem to suggest to do simplification, so I also did but don't know what went wrong. :( )
Here's what the Book's explanation says which i understand very well but what's wrong with the above said approach is what I am trying to understand,
