Question

rewrite the expression completely.
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perform multiplication
\\(dfrac{w^2 - 9}{w + 3} cdot dfrac{w - 3}{w^2 - 9}\\)
enter the result.

Explanation:

Step1: Factor the difference of squares

Notice that $w^2 - 9 = (w+3)(w-3)$

Step2: Substitute factored form into expression

$$\frac{(w+3)(w-3)}{w+3} \cdot \frac{w-3}{(w+3)(w-3)}$$

Step3: Cancel common factors

Cancel $(w+3)$ from first fraction, cancel $(w-3)$ across fractions:
$$(w-3) \cdot \frac{1}{w+3}$$

Step4: Multiply remaining terms

$$\frac{w-3}{w+3}$$
(Note: $w
eq -3, 3$ to avoid division by zero)

Answer:

$\frac{w-3}{w+3}$