Question 53255
First flip the second fraction over and multiply:
{{{((x^3-1)/(x^2+1))*((x^2-x)/(9x^2+9x+9))}}}
Then factor. The numerator in the first fraction is a difference between two cubes can be factored:
{{{a^3-b^3=(a-b)(a^2+ab+b^2)}}} 
in your case a=x and b=1.

{{{((x-1)(x^2+x+1)/(x^2+1))*(x(x-1)/(9(x^2+x+1)))}}}
{{{(x(x-1)^2(x^2+x+1))/(9(x^2+1)(x^2+x+1))}}}
{{{x^2+x+1}}}cancels and we have:
{{{(x(x-1)^2)/(9(x^2+1))}}}