What is the product? \( \frac{3 k}{k+1} \cdot \frac{k^{2}-1}{3 k^{3}} \)

1. Start with the product: \[ \frac{3k}{k+1} \cdot \frac{k^{2}-1}{3k^{3}} \] 2. Factor the quadratic in the numerator: \[ k^2 - 1 = (k-1)(k+1) \] 3. Substitute the factorization into the expression: \[ \frac{3k}{k+1} \cdot \frac{(k-1)(k+1)}{3k^{3}} \] 4. Cancel the common factor \(3\) in the numerator and denominator: \[ \frac{k}{k+1} \cdot \frac{(k-1)(k+1)}{k^{3}} \] 5. Cancel the common factor \((k+1)\): \[ \frac{k}{\cancel{k+1}} \cdot \frac{(k-1)\,\cancel{k+1}}{k^{3}} = \frac{k(k-1)}{k^{3}} \] 6. Cancel the common factor \(k\): \[ \frac{\cancel{k}(k-1)}{\cancel{k}k^{2}} = \frac{k-1}{k^{2}} \] The final simplified product is: \[ \frac{k-1}{k^{2}} \]