what is the product?\n$\frac{3k}{k + 1}cdot\frac{k^{2}-1}{3k^{3}}$\n$\frac{k + 1}{k^{2}}$\n$\frac{k…
what is the product?\n$\frac{3k}{k + 1}cdot\frac{k^{2}-1}{3k^{3}}$\n$\frac{k + 1}{k^{2}}$\n$\frac{k - 1}{k^{2}}$\n$\frac{-1}{k}$\n$\frac{1}{k}$
Answer
Answer:
B. $\frac{k - 1}{k^{2}}$
Explanation:
Step1: Factor the numerator $k^{2}-1$
$k^{2}-1=(k + 1)(k - 1)$
Step2: Multiply the fractions
$\frac{3k}{k + 1}\cdot\frac{k^{2}-1}{3k^{3}}=\frac{3k(k + 1)(k - 1)}{(k + 1)\cdot3k^{3}}$
Step3: Cancel out common factors
Cancel out the common factors $3k$ and $(k + 1)$ in the numerator and denominator. We get $\frac{k - 1}{k^{2}}$