Simplify (3k)/(k+1)*(k^2-1)/(3k^3)

$\frac{3 k}{k + 1} \cdot \frac{k^{2} - 1}{3 k^{3}}$

Step 1

Simplify the numerator.

Rewrite $1$ as $1^{2}$ .

$\frac{3 k}{k + 1} \cdot \frac{k^{2} - 1^{2}}{3 k^{3}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = k$ and $b = 1$ .

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

Step 2

Reduce the expression by cancelling the common factors.

Cancel the common factor of $3 k$ .

Factor $3 k$ out of $3 k^{3}$ .

$\frac{3 k}{k + 1} \cdot \frac{(k + 1) (k - 1)}{3 k (k^{2})}$

Cancel the common factor.

$\frac{3 k}{k + 1} \cdot \frac{(k + 1) (k - 1)}{3 k \cdot k^{2}}$

Rewrite the expression.

$\frac{1}{k + 1} \cdot \frac{(k + 1) (k - 1)}{k^{2}}$

Cancel the common factor of $k + 1$ .

Cancel the common factor.

$\frac{1}{k + 1} \cdot \frac{(k + 1) (k - 1)}{k^{2}}$

Rewrite the expression.

$\frac{k - 1}{k^{2}}$