Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[\frac{\frac{3x\times 1}{(x-1)(x+1)}}{\frac{3}{{x}^{2}-1}}\]
Simplify \(3x\times 1\) to \(3x\).
\[\frac{\frac{3x}{(x-1)(x+1)}}{\frac{3}{{x}^{2}-1}}\]
Rewrite \({x}^{2}-1\) in the form \({a}^{2}-{b}^{2}\), where \(a=x\) and \(b=1\).
\[\frac{\frac{3x}{(x-1)(x+1)}}{\frac{3}{{x}^{2}-{1}^{2}}}\]
Use Difference of Squares: \({a}^{2}-{b}^{2}=(a+b)(a-b)\).
\[\frac{\frac{3x}{(x-1)(x+1)}}{\frac{3}{(x+1)(x-1)}}\]
Simplify.
\[\frac{\frac{3x}{(x+1)(x-1)}}{\frac{3}{(x+1)(x-1)}}\]
Cancel denominators.
\[\frac{3x}{3}\]
Cancel \(3\).
\[x\]
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