How to solve $(z^1+z^2+z^3+z^4)^3$ using Pascals Triangle? In an exercise it seems I must use Pascal's triangle to solve this $(z^1+z^2+z^3+z^4)^3$. The result would be $z^3 + 3z^4 + 6z^5 + 10z^ 6 + 12z^ 7 + 12z^ 8 + 10z^ 9 + 6z^ {10} + 3z^ {11} + z^{12}$. But how do I use the triangle to get to that result? Personally I can only solve things like $(x+y)^2$ and $(x+y)^3$.
Thanks for any tips that may be given.
 A: $$(z+z^2+z^3+z^4)^3 = z^3\cdot\left(\frac{1-z^4}{1-z}\right)^3=z^3(1-3z^4+3z^8-z^{12})\sum_{n\geq 0}\binom{n+2}{2}z^n$$
hence:
$$\begin{eqnarray*}(z+z^2+z^3+z^4)^3&=&(z^3-3z^7+3z^{11}-z^{15})\sum_{n\geq 0}\binom{n+2}{2}z^n\\&=&\sum_{n\geq 3}\binom{n-1}{2}z^n-3\sum_{n\geq 7}\binom{n-5}{2}+3\sum_{n\geq 11}\binom{n-9}{2}z^n-\sum_{n\geq 15}\binom{n-13}{2}z^n\\&=&\sum_{n= 3}^{12}\binom{n-1}{2}z^n-3\sum_{n=7}^{12}\binom{n-5}{2}z^n+3\sum_{n=11}^{12}\binom{n-9}{2}z^n\\&=&\frac{1}{2}\left(\sum_{n= 3}^{6}(n-1)(n-2)z^n-2\sum_{n=7}^{10}(n-4)(n-11)z^n+\sum_{n=11}^{12}(n-13)(n-14)z^n\right)\end{eqnarray*}$$
A: You can have x = z1
and y = z 2 + z 3 + z 4
Then the problem is (x+y) 3. Later resubstitute y in the expansion and choose a = z 2 and b = z 3 + z 4. And expand those powers by pascal's triangle and finally resubstitute for a and b and expand those powers by pascal's triangle.
A: The expression factors as
$$z^3(1+z)^3(1+z^2)^3=z^3(1+3z+3z^2+z^3)(1+3z^2+3z^4+z^6).$$
Then I see no better way than to perform the multiply (though there is a symmetry)
$$\begin{align}
&1+3z+&3z^2+&z^3\\
&&3z^2+&9z^3+&9z^4+&3z^5\\
&&&&3z^4+&9z^5+&9z^6+&3z^7\\
&&&&&&z^6+&3z^7+&3z^8+z^9\\
\end{align}$$
