The question asks us to express $64 = a^b$ in four different ways, where $a$ and $b$ are positive integers. We need to find all such pairs of $a$ and $b$, and then determine the sum of all possible values of $a$.

Step 1: Find all representations of $64 = a^b$

We know that $64 = 2^6$. Let's break down all the possible ways to express 64 as $a^b$:

1. $64 = 64^1$ → Here, $a = 64$ and $b = 1$.
2. $64 = 8^2$ → Here, $a = 8$ and $b = 2$, since $8^2 = (2^3)^2 = 2^6$.
3. $64 = 4^3$ → Here, $a = 4$ and $b = 3$, since $4^3 = (2^2)^3 = 2^6$.
4. $64 = 2^6$ → Here, $a = 2$ and $b = 6$.

Step 2: Sum of all values of $a$

The possible values of $a$ are $64, 8, 4,$ and $2$.

So, the sum of these values is: $64 + 8 + 4 + 2 = 78$

Final Answer:

The sum of all possible values of $a$ is 78.

Would you like further details on how I derived these steps?

Here are some related questions:

1. What other integers can be expressed as powers in multiple ways like 64?
2. Can the values of $b$ for such equations ever be non-integers?
3. How would the solution change if negative integers were allowed for $a$?
4. What if the question asked for real numbers instead of positive integers?
5. How does the prime factorization of numbers help in solving these types of problems?

Tip: When dealing with powers, always check if the base can be broken down into smaller powers of prime numbers for better clarity!