<PRE><B>Question 768707</B>
Factor

x&lt;sup&gt;4&lt;/sup&gt; + 324
&lt;pre&gt;
(x&lt;sup&gt;2&lt;/sup&gt;)&lt;sup&gt;2&lt;/sup&gt; + 18&lt;sup&gt;2&lt;/sup&gt;

That would be a perfect square if the term that is
twice the product of the square roots of those two
terms were added.  That&#39;s 2·18·x&lt;sup&gt;2&lt;/sup&gt; or 36x&lt;sup&gt;2&lt;/sup&gt;.

So we add that between those terms and then subtract it.
That&#39;s the same as adding 0:

(x&lt;sup&gt;2&lt;/sup&gt;)&lt;sup&gt;2&lt;/sup&gt; + 36x&lt;sup&gt;2&lt;/sup&gt; + 18&lt;sup&gt;2&lt;/sup&gt; - 36x&lt;sup&gt;2&lt;/sup&gt;

Factor the first three terms as (x&lt;sup&gt;2&lt;/sup&gt; + 18)(x&lt;sup&gt;2&lt;/sup&gt; + 18) or (x&lt;sup&gt;2&lt;/sup&gt; + 18)&lt;sup&gt;2&lt;/sup&gt;

(x&lt;sup&gt;2&lt;/sup&gt; + 18)&lt;sup&gt;2&lt;/sup&gt; - 36x&lt;sup&gt;2&lt;/sup&gt;

That is the difference of two perfect squares:

[(x&lt;sup&gt;2&lt;/sup&gt; + 18) - 6x][(x&lt;sup&gt;2&lt;/sup&gt; + 18) + 6x]

[x&lt;sup&gt;2&lt;/sup&gt; + 18 - 6x][x&lt;sup&gt;2&lt;/sup&gt; + 18 + 6x]

(x&lt;sup&gt;2&lt;/sup&gt; - 6x + 18)(x&lt;sup&gt;2&lt;/sup&gt; + 6x + 18)

Edwin&lt;/pre&gt;</PRE>