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

To find the product \( \frac{3 k}{k+1} \cdot \frac{k^{2}-1}{3 k^{3}} \), we can start by multiplying the numerators and the denominators. The numerator is: \[ 3k \cdot (k^2 - 1) = 3k(k^2 - 1) = 3k(k - 1)(k + 1) \] The denominator is: \[ (k + 1) \cdot 3k^3 = 3k^3(k + 1) \] Now we can express the product as: \[ \frac{3k(k - 1)(k + 1)}{3k^3(k + 1)} \] Next, we can cancel \( 3 \) from the numerator and denominator: \[ \frac{k(k - 1)(k + 1)}{k^3(k + 1)} \] Next, we can cancel \( k + 1 \) from the numerator and the denominator (as long as \( k \neq -1 \)): \[ \frac{k(k - 1)}{k^3} \] Finally, we can simplify it further: \[ \frac{k - 1}{k^2} \] So the product is: \[ \frac{k - 1}{k^2}, \quad (k \neq -1) \]