Correct. You recognized that $\displaystyle\frac{{{6}{{x}}^{{3}}-{6}{x}}}{{{3}{{x}}^{{3}}-{6}{{x}}^{{2}}-{9}{x}}}=\frac{{{6}{x}{\left({x}-{1}\right)}{\left({x}+{1}\right)}}}{{{3}{x}{\left({x}-{3}\right)}{\left({x}+{1}\right)}}}\usepackage{cmbright}\Huge\$ . Pulling out factors of 1, you have $\displaystyle\frac{{{\left({3}{x}\right)}{\left({x}+{1}\right)}}}{{{\left({3}{x}\right)}{\left({x}+{1}\right)}}}={1}\usepackage{cmbright}\Huge\$ , so $\displaystyle\frac{{{6}{x}{\left({x}-{1}\right)}{\left({x}+{1}\right)}}}{{{3}{x}{\left({x}-{3}\right)}{\left({x}+{1}\right)}}}={1}\cdot\frac{{{2}{\left({x}-{1}\right)}}}{{{x}-{3}}}=\frac{{{2}{\left({x}-{1}\right)}}}{{{x}-{3}}}\usepackage{cmbright}\Huge\$ .