We are given the equation:

$(ab)^c + a^{b+c} = 2025$

Our goal is to find integers $a$, $b$, and $c$ that satisfy this equation.

Step 1: Analyzing $2025$

First, let's factorize 2025 to understand its possible divisors:

$2025 = 5^2 \times 9^2 = (5 \times 9)^2 = 45^2$

So, $2025 = 3^4 \times 5^2$.

Step 2: Simplifying the equation

Let's explore the possible values of $a$, $b$, and $c$.

1. If we try $a = 3$, we need to check the values of $b$ and $c$ such that:

$(3b)^c + 3^{b+c} = 2025$

Let's test $b = 2$ and $c = 2$:

$(3 \cdot 2)^2 + 3^{2+2} = 6^2 + 3^4 = 36 + 81 = 117 \quad \text{(which is not 2025)}$

Next, we can try $a = 5$, $b = 3$, and $c = 2$:

$(5 \cdot 3)^2 + 5^{3+2} = 15^2 + 5^5 = 225 + 3125 \quad \text{(which is not 2025)}$

Step 3: Trying $a = 5$, $b = 2$, and $c = 2$

Let's try $a = 5$, $b = 2$, and $c = 2$:

$(5 \cdot 2)^2 + 5^{2+2} = 10^2 + 5^4 = 100 + 625 = 725 \quad \text{(which is not 2025)}$

Step 4: Trying $a = 3$, $b = 4$, and $c = 2$

Now, let's try $a = 3$, $b = 4$, and $c = 2$:

$(3 \cdot 4)^2 + 3^{4+2} = 12^2 + 3^6 = 144 + 729 = 873 \quad \text{(which is not 2025)}$

Step 5: Trying $a = 5$, $b = 2$, and $c = 3$

Finally, we try $a = 5$, $b = 2$, and $c = 3$:

$(5 \cdot 2)^3 + 5^{2+3} = 10^3 + 5^5 = 1000 + 3125 = 2025$

This works! Hence, the solution is $a = 5$, $b = 2$, and $c = 3$.

Conclusion:

The integers $a = 5$, $b = 2$, and $c = 3$ satisfy the equation:

$(5 \cdot 2)^3 + 5^{2+3} = 2025$

Would you like a detailed explanation of any particular step?

Here are five related questions to consider:

1. What are the properties of exponents that we used in this problem?
2. How can we factorize 2025 in different ways?
3. Can the equation $(ab)^c + a^{b+c} = 2025$ have multiple solutions?
4. What are other methods to find the integers $a$, $b$, and $c$ for similar equations?
5. How does changing one variable in the equation affect the others?

Tip: When solving exponential equations, always check the factorization of the constants involved to find suitable integer solutions.