Problem 494

Solution. In the numerator and denominator of each fraction, we factor out the common factor:

$$\frac{3 a^{2}+3 a b+3 b^{2}}{4 a+4 b} \cdot \frac{2 a^{2}-2 b^{2}}{9 a^{3}-9 b^{3}}=\frac{3\left(a^{2}+a b+b^{2}\right)}{4(a+b)} \cdot \frac{2\left(a^{2}-b^{2}\right)}{9\left(a^{3}-b^{3}\right)}$$

Using the formulas for the difference of squares and the difference of cubes, we get

$$\frac{3\left(a^{2}+a b+b^{2}\right) \cdot 2(a-b)(a+b)}{4(a+b) \cdot 9(a-b)\left(a^{2}+a b+b^{2}\right)}=\frac{3 \cdot 2}{4 \cdot 9}=\frac{1}{6}$$