Question

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

Explanation:

Step1: Factor the quadratic term

$k^2 - 1 = (k+1)(k-1)$

Step2: Rewrite the product

$\frac{3k}{k+1} \cdot \frac{(k+1)(k-1)}{3k^3}$

Step3: Cancel common factors

Cancel $3k$ and $k+1$: $\frac{k-1}{k^2}$

Answer:

$\boldsymbol{\frac{k-1}{k^2}}$ (corresponding to the second option)