<PRE><B>Question 1092206</B>
&lt;br&gt;I&#39;ll show you a couple of these so you can get the idea of how to work them.  Then you can try the others yourself.&lt;br&gt;
The last question is the easiest one to use to show you the general concept for working this kind of problem.  In that problem we are given&lt;br&gt;
{{{x + 1/x = 4}}}&lt;br&gt;
and we are to show that&lt;br&gt;
{{{x^2 + 1/x^2 = 14}}}&lt;br&gt;
Observe what happens when you square both sides of the given equation.  When you square the binomial on the left, the middle term is a constant.  That is the general principle that lets you solve problems like this.&lt;br&gt;
{{{(x + 1/x)^2 = 4^2}}}
{{{x^2 + 2 + 1/x^2 = 16}}}
{{{x^2 + 1/x^2 = 14}}}&lt;br&gt;
Now let&#39;s look a bit at your second problem.  There you are given&lt;br&gt;
{{{a+1/a = 7}}}&lt;br&gt;
and you are asked to find the values of&lt;br&gt;
1. a^2+1/a^2
2.(a-1/a)^2
3. a^4+1/a^4&lt;br&gt;
Finding the first of these will be exactly like the example worked above.  For the second one, note that&lt;br&gt;
{{{(a-1/a)^2 = a^2-2+1/a^2}}}&lt;br&gt;
so that one will be easy after you have solved the first one.  And for the third of these, you can get the answer in a similar fashion by squaring your answer for the first one.</PRE>