Question 66380
<pre><font size = 5 color = "darkblue"><b>
      x<sup>4</sup> - 1
------------------
 x<sup>3</sup> + x<sup>2</sup> + x + 1

Factor the numerator completely:

                x<sup>4</sup> - 1

That's the difference of two perfect squares, so

             (x<sup>2</sup> - 1)(x<sup>2</sup> + 1)

The first factor is also the difference of two
perfect squares, so the complete factorization is:

         (x - 1)(x + 1)(x<sup>2</sup> + 1)

(Note that you CANNOT factor the SUM of two perfect
 squares, only the DIFFERENCE of two perfect squares)

Factor the denominator completely, by grouping:

 x<sup>3</sup> + x<sup>2</sup> + x + 1

Factor x<sup>2</sup> out of the first two terms:

 x<sup>2</sup>(x + 1) + x + 1

Factor 1 out of the last two terms:

 x<sup>2</sup>(x + 1) + 1(x + 1)

Now factor (x + 1) out of both terms:

 (x + 1)(x<sup>2</sup> + 1)

Now return to the original expression:

      x<sup>4</sup> - 1
------------------
 x<sup>3</sup> + x<sup>2</sup> + x + 1

and replace the numerator and denominator
by their complete factorizations:

  (x - 1)(x + 1)(x<sup>2</sup> + 1)  
-------------------------
    (x + 1)(x<sup>2</sup> + 1)

Cancel the (x + 1)'s and the (x<sup>2</sup> + 1)'s

            1       1
  (x - 1)(<s>x + 1</s>)(<s>x² + 1</s>)  
-------------------------
    (<s>x + 1</s>)(<s>x² + 1</s>)
       1       1

and all you have left is a measly:

      x - 1
 
Edwin</pre>