Question 625860
<pre>

 x-1       x
————— + ——————
  x³     x²+1

The LCD is x³(x²+1), so we multiply each
fraction by whichever factor of the LCD its
denominator is lacking, over itself.

The first fraction's denominator lacks the 
factor x²+1 and the second fraction's 
denominator lacks the factor x³, so we put 
the missing factors over themselves and 
multiply, so that the denominators will both
become the LCD.

 x-1     x²+1      x       x³
————— · —————— + —————— · ————
 x³      x²+1     x²+1     x³



(x-1)(x²+1)     (x)(x³)
——————————— + ———————————— 
  x³(x²+1)     (x²+1)(x³)

 x³+x-x²-1        x<sup>4</sup>
——————————— + ———————————— 
  x³(x²+1)     (x²+1)(x³)

 x³+x-x²-1+x<sup>4</sup>
—————————————— 
   x³(x²+1) 

 x<sup>4</sup>+x³-x²+x-1
—————————————— 
   x³(x²+1)

Edwin</pre>