We are given the equation:

$a + \frac{1}{a + 2} = 3$

Step 1: Solve for $a$

First, subtract $a$ from both sides of the equation:

$\frac{1}{a + 2} = 3 - a$

Next, take the reciprocal of both sides to isolate $a + 2$:

$a + 2 = \frac{1}{3 - a}$

Multiply both sides by $(3 - a)$ to eliminate the denominator:

$(a + 2)(3 - a) = 1$

Now, expand the left-hand side:

$3a + 6 - a^2 - 2a = 1$

Simplify:

$a + 6 - a^2 = 1$

Rearrange the equation:

$-a^2 + a + 6 = 1$

Move all terms to one side:

$-a^2 + a + 5 = 0$

Multiply by $-1$ to simplify:

$a^2 - a - 5 = 0$

Now, solve this quadratic equation using the quadratic formula:

$a = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-5)}}{2(1)}$

Simplify:

$a = \frac{1 \pm \sqrt{1 + 20}}{2}$

$a = \frac{1 \pm \sqrt{21}}{2}$

Thus, the two possible values for $a$ are:

$a = \frac{1 + \sqrt{21}}{2} \quad \text{or} \quad a = \frac{1 - \sqrt{21}}{2}$

Step 2: Solve for $(a + 2)^5 - \frac{1}{(a + 2)^5}$

Recall from earlier that:

$a + 2 = \frac{1}{3 - a}$

Now, raise both sides of this equation to the 5th power:

$(a + 2)^5 = \left( \frac{1}{3 - a} \right)^5$

So, we need to find:

$(a + 2)^5 - \frac{1}{(a + 2)^5}$

Since $(a + 2)^5 = \left( \frac{1}{3 - a} \right)^5$, this expression simplifies to:

$\boxed{0}$

Final Answer:

$(a + 2)^5 - \frac{1}{(a + 2)^5} = 0$

Would you like more details or have any questions?

Here are 5 related questions to expand on the concept:

1. How do you solve quadratic equations using factorization?
2. What is the significance of the quadratic formula?
3. Can you solve a cubic equation using a similar approach?
4. How do reciprocal functions behave in different equations?
5. How would the solution change if $a + \frac{1}{a+2} = 4$?

Tip: Always check if equations can be simplified before solving to reduce complexity!