Question 390385
<pre>
Add:
{{{3/(x^3-x)}}}{{{""+""}}}{{{4/(x^2+2x+1)}}}

Factor the first denominator completely:

{{{x^3-x}}}{{{""=""}}}{{{x(x^2-1)}}}{{{""=""}}}{{{x(x-1)(x+1)}}}

Factor the second denominator completely:

{{{x^2+2x+1}}}{{{""=""}}}{{{(x+1)(x+1)}}}{{{""=""}}}{{{(x+1)^2}}}

{{{3/(x(x-1)(x+1))}}}{{{""+""}}}{{{4/(x+1)^2}}}

Creating the LCD:

Rule:
The LCD must contain each factor of each denominator to the largest power
it appears in any denominator.

The factor {{{x}}} occurs to the first power in the first fraction,
and not at all in the second fraction, so the LCD will contain {{{x}}}
to the first power.

The factor {{{(x-1)}}} occurs to the first power in the first fraction,
and not at all in the second fraction, so the LCD will contain {{{(x-1)}}}
to the first power.

The factor {{{(x+1)}}} occurs to the first power in the first fraction,
and to the 2nd power in the second fraction, so the LCD will contain 
{{{(x+1)}}} to the 2nd power, or {{{(x+1)^2}}}

Therefore the LCD = {{{x(x-1)(x+1)^2}}}

{{{3/(x(x-1)(x+1))}}}{{{""+""}}}{{{4/(x+1)^2}}}

The first fraction needs to be multiplied by {{{red(((x+1))/((x+1)))}}} so that
its denominator will become the LCD:

{{{
expr(3/(x(x-1)(x+1)))red(((x+1))/((x+1)))
}}}{{{""+""}}}{{{4/(x+1)^2}}}


The second fraction needs to be multiplied by {{{red(  (x(x-1))/(x(x-1))  )}}}
so that its denominator will also become the LCD:



{{{
expr(3/(x(x-1)(x+1)))red(((x+1))/((x+1)))
}}}{{{""+""}}}{{{  
expr(red(  (x(x-1))/(x(x-1))  ))*
expr(4/(x+1)^2)
}}}


{{{  (3(x+1))/(x(x-1)(x+1)^2)}}}{{{""+""}}}{{{(4x(x-1))/(x(x-1)(x+1)^2)}}}

Multiply the numerators out but do not multiply the
denominators out.

{{{  (3x+3)/(x(x-1)(x+1)^2)}}}{{{""+""}}}{{{(4x^2-4x)/(x(x-1)(x+1)^2)}}}

Combine the numerators over the LCD:

{{{(3x+3+4x^2-4x)/(x(x-1)(x+1)^2)}}}

{{{(4x^2-x+3)/(x(x-1)(x+1)^2)}}}

The numerator doesn't factor, so we leave it like that.

Edwin</pre>