Higher order terms are neglected and only $x^3$ and $x^9$ terms are useful.
you explained very well! much obliged!!
Please let me know how how did you reach this equation
and why are you loking for the coefficient of x^12, please provide details
it is same like distribution of indistinguishable balls into distinguishable boxes.
Say where x1+x2+x3=12
x1,x2,x3 can have value between 1,2,3,4,5,6.
Each have the same (x+x^2+x^3+x^4+x^5+x^6) so taken cube ....and n=12 so finding 12th term
Can you please explain this part from your solution:
find coeff of x12 in (x^6+x^5+x^4+x^3+x^2+x)3
In d link mentioned below. Yeah. :)