The given expression is:

$\frac{a + 3}{a^3 - 4a} \cdot \frac{a^2 + 4a + 4}{5a + 15} \cdot \frac{2 + a}{a^2 - 2a}.$

Step-by-step solution:

Step 1: Factorize all terms

1. First fraction:

   • Denominator $a^3 - 4a$: $a^3 - 4a = a(a^2 - 4) = a(a - 2)(a + 2).$ So, the fraction becomes: $\frac{a + 3}{a(a - 2)(a + 2)}.$

2. Second fraction:

   • Numerator $a^2 + 4a + 4$: $a^2 + 4a + 4 = (a + 2)^2.$
   • Denominator $5a + 15$: $5a + 15 = 5(a + 3).$ So, the fraction becomes: $\frac{(a + 2)^2}{5(a + 3)}.$

3. Third fraction:

   • Numerator $2 + a$: Since $2 + a = a + 2$, the numerator remains $a + 2$.
   • Denominator $a^2 - 2a$: $a^2 - 2a = a(a - 2).$ So, the fraction becomes: $\frac{a + 2}{a(a - 2)}.$

Step 2: Combine the fractions

The expression becomes: $\frac{a + 3}{a(a - 2)(a + 2)} \cdot \frac{(a + 2)^2}{5(a + 3)} \cdot \frac{a + 2}{a(a - 2)}.$

Step 3: Cancel common factors

1. $a + 3$ in the numerator of the first fraction cancels with $a + 3$ in the denominator of the second fraction.
2. $a + 2$ appears three times in the numerator and once in the denominator, so two $a + 2$'s remain in the numerator.
3. $a$ appears in the denominator of the first and third fractions, so $a^2$ remains in the denominator.
4. $a - 2$ appears in the denominator of the first and third fractions, so $a - 2$^2 remains in the denominator.

After cancellation, we are left with: $\frac{(a + 2)^2}{5a^2(a - 2)^2}.$

Final Answer:

$\boxed{\frac{(a + 2)^2}{5a^2(a - 2)^2}}$

Would you like a more detailed breakdown of any step or have additional questions? 😊

Related Questions:

1. How do you simplify fractions involving polynomials?
2. What are the key techniques for factoring cubic expressions?
3. Why is it important to cancel common terms in a fraction?
4. Can you show another example of combining and simplifying rational expressions?
5. How would the answer change if a specific value of $a$ is substituted?

Tip:

Always check for restrictions on the variable $a$. For this problem, $a \neq 0, \pm 2, -3$ to avoid division by zero.