(x^2 -1) / 3  * 2/(x^2 - x)   let's "swap' denominators

(x^2 -1) / (x^2 -x) * 2/3    factoring the fraction on the left on top and bottom, we have

[ (x -1)(x + 1)] / [x (x -1)] *2/3       note.... the (x-1) terms "cancel"...and we're left with

(x + 1)/x  *  2/3 =

[2(x+ 1)] / [3x)  =  (2x + 2) / (3x)

Whenever you multiply fractions, just multiply the denominator by the denominator and the numerator by the numerator: (x^2 - 1) * (2) = 2x^2 - 2 and 3 * (x^2 - x) = 2x^2 - 2x, so therefore:

(x^2-1)/3 * (2)/(x^2-x) = (2x^2 - 2)/(2x^2 - 2x)

The 2x^2's cancel eachother out and so do the -2's. So, your simplified expression is 1/x.

@Mathematician

Can you explain it more simple?

Sorry