1. \( \frac{a+2}{a^{2}+2 a+4} \cdot \frac{a^{3}-8}{2 a^{2}-8} \)

To simplify the expression \( \frac{a+2}{a^{2}+2a+4} \cdot \frac{a^{3}-8}{2a^{2}-8} \), let's break it down step by step. First, factor each part of the expression. 1. **Numerator of the first fraction**: \( a + 2 \) remains as it is. 2. **Denominator of the first fraction**: \( a^2 + 2a + 4 \) can be checked for factoring, but it does not factor nicely over the reals. 3. **Numerator of the second fraction**: \( a^3 - 8 \) is a difference of cubes, which can be factored as: \[ a^3 - 8 = (a - 2)(a^2 + 2a + 4) \] 4. **Denominator of the second fraction**: \( 2a^2 - 8 \) can be factored as: \[ 2a^2 - 8 = 2(a^2 - 4) = 2(a - 2)(a + 2) \] Now substituting back into the expression, we have: \[ \frac{a + 2}{a^2 + 2a + 4} \cdot \frac{(a - 2)(a^2 + 2a + 4)}{2(a - 2)(a + 2)} \] Next, we can cancel common factors. Notice that \( a^2 + 2a + 4 \) in the numerator of the second fraction cancels with the denominator of the first fraction, resulting in: \[ \frac{a + 2}{1} \cdot \frac{(a - 2)}{2(a - 2)(a + 2)} \] Now we can also cancel \( (a + 2) \) and \( (a - 2) \) (under the condition that \( a \neq -2 \) and \( a \neq 2 \)): \[ \frac{1}{2} \] So, the simplified expression is: \[ \frac{1}{2} \] This simplification holds true for all \( a \) except when \( a = -2 \) and \( a = 2\) (where the original expression is undefined).