Expansion of polynomial raised to high power Is there an easy way to expand something like (x + x^2 + x^3)^6 ?
Thanks in advance!
 A: My approach avoids multinomials:
Let's play with it a bit:
$$\begin{align}
(x+x^2+x^3)^6 &= x^6(1+x+x^2)^6\\
&= x^6\frac{((1-x)(1+x+x^2))^6}{(1-x)^6}\\
&= \frac{x^6(1-x^3)^6}{(1-x)^6} \\
\end{align}$$
The numerator and denominator can now be expanded with the binomial theorem, and then long division will simplify the expression. (This is the way I would do it.)
Another approach: 
$$\begin{align}
(x+x^2+x^3)^6 &= (x+(x^2+x^3))^6 \\
&= x^6+6x^5(x^2+x^3) + 15x^4(x^2+x^3)^2 \\
& +\, 20x^3(x^2+x^3)^3 + 15x^2(x^2+x^3)^4 + 6x(x^2+x^3)^5 + (x^2+x^3)^6
\end{align}$$
Now, each of the binomials can be expanded as above.
A: There is a very compact solution for expanding an all-ones polynomial with no constant term presented here, demonstrated with SymPy/Python below:
>>> from sympy import binomial
>>> Limits = lambda i,j: range(i,j+1)  # to give values i-j
>>> list(Limits(1,3))
>>> [1, 2, 3]

>>> k=3  # for x + x**2 + x**3
>>> n=6
>>> [sum((-1)**i*binomial(n,i)*binomial(r-i*k-1,n-1)
...  for i in Limits(0,(r-n)//k)) for r in Limits(n,k*n)]
[1, 6, 21, 50, 90, 126, 141, 126, 90, 50, 21, 6, 1]

In your case having row 6 and diagonal 5 (down through row 13) of Pascal's triangle will help with the binomials.
