Okay, I have the expression:

$\displaystyle

\frac{6}{2(x - 1)(x + 1)} + \frac{1}{6(x - 1)^2}

$

now I need help to understand how these fractions are added together.

I have the solution to it, but upon analysing it I am a bit confused:

$\displaystyle

\frac{6}{2(x - 1)(x + 1)} + \frac{1}{6(x - 1)^2}

$

$\displaystyle

= \frac{6}{2(x - 1)(x + 1)} \times \frac{3(x - 1)}{3(x - 1)}

+ \frac{1}{6(x - 1)^2} \times \frac{x + 1}{x + 1} \\

$

$\displaystyle

= \frac{18(x - 1)}{6(x - 1)^2(x + 1)} +

\frac{x + 1}{6(x - 1)^2(x + 1)}

$

$\displaystyle

= \frac{18(x - 1) + x + 1}{6(x - 1)^2(x + 1)}

$

- Why are we multiplying $\displaystyle \frac{6}{2(x - 1)(x + 1)}$ by $\displaystyle \frac{3(x - 1)}{3(x - 1)} $ and how did we get the conclusion to use $\displaystyle 3(x - 1)$ in the first place.
- Same with $\displaystyle \frac{1}{6(x - 1)^2}$ and $\displaystyle x + 1$.

I get the rest of it but those two parts of the solution don't make sense to me!?

Are they using an Least Common Denominator (LCD) or something?

Thanks in advance.