<PRE><B>Question 1031691</B>
&lt;pre&gt;
{{{(x^4 + x^3 + x^2 + x + 1)^4}}}{{{&quot;&quot;=&quot;&quot;}}}

{{{(x^4 + x^3 + x^2 + x + 1)(x^4 + x^3 + x^2 + x + 1)(x^4 + x^3 + x^2 + x + 1)(x^4 + x^3 + x^2 + x + 1)}}}

The only way to get a x in that multiplication is: 

1. When the 1&#39;s in the 1st, 2nd, and 3rd parentheses are multiplied 
   by the x in the 4th parentheses, giving term 1*1*1*x = x.

2. When the 1&#39;s in the 1st, 2nd, and 4th parentheses are multiplied 
   by the x in the 3rd parentheses, giving term 1*1*x*1 = x.

3. When the 1&#39;s in the 1st, 3rd, and 4th parentheses are multiplied 
   by the x in the 2nd parentheses, giving term 1*x*1*1 = x.

4. When the 1&#39;s in the 2nd, 3rd, and 4th parentheses are multiplied 
   by the x in the 1st parentheses, giving term x*1*1*1 = x.

So the only term in x is x+x+x+x = 4x, which has coefficient 4.

Incidentally, the whole thing multiplied out is:
&lt;font size=1&gt;
x&lt;sup&gt;16&lt;/sup&gt;+4x&lt;sup&gt;15&lt;/sup&gt;+10x&lt;sup&gt;14&lt;/sup&gt;+20x&lt;sup&gt;13&lt;/sup&gt;+35x&lt;sup&gt;12&lt;/sup&gt;+52x&lt;sup&gt;11&lt;/sup&gt;+68x&lt;sup&gt;10&lt;/sup&gt;+80x&lt;sup&gt;9&lt;/sup&gt;+85x&lt;sup&gt;8&lt;/sup&gt;+80x&lt;sup&gt;7&lt;/sup&gt;+68x&lt;sup&gt;6&lt;/sup&gt;+52x&lt;sup&gt;5&lt;/sup&gt;+35x&lt;sup&gt;4&lt;/sup&gt;+20x&lt;sup&gt;3&lt;/sup&gt;+10x&lt;sup&gt;2&lt;/sup&gt;+4x+1&lt;/font&gt;

Edwin&lt;/pre&gt;</PRE>