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3. $\frac{a^{2}-b^{2}}{3a + 3b}cdot\frac{12}{4a - 4b}$

Question

  1. $\frac{a^{2}-b^{2}}{3a + 3b}cdot\frac{12}{4a - 4b}$

Explanation:

Step1: Factor the expressions

We know that \(a^{2}-b^{2}=(a + b)(a - b)\), \(3a + 3b=3(a + b)\) and \(4a-4b = 4(a - b)\). So the original - expression \(\frac{a^{2}-b^{2}}{3a + 3b}\cdot\frac{12}{4a - 4b}\) becomes \(\frac{(a + b)(a - b)}{3(a + b)}\cdot\frac{12}{4(a - b)}\).

Step2: Cancel out the common factors

Cancel out the common factors \((a + b)\) and \((a - b)\) in the numerator and denominator. We get \(\frac{1}{3}\cdot\frac{12}{4}\).

Step3: Simplify the remaining fraction

\(\frac{12}{3\times4}=\frac{12}{12}=1\).

Answer:

1