Let $a, b, c$ be positive integers such that
$$\gcd(a, b) + \text{lcm}(a, b) = \gcd(a, c) + \text{lcm}(a, c).$$
Does it follow from this that $b = c$?
1. Given the equation:
   \[
   \gcd(a, b) + \text{lcm}(a, b) = \gcd(a, c) + \text{lcm}(a, c)
   \]
   we need to determine if \( b = c \) follows from this.

2. Recall the relationship between the greatest common divisor (gcd) and the least common multiple (lcm):
   \[
   \text{lcm}(x, y) = \frac{x \cdot y}{\gcd(x, y)}
   \]

3. Substitute the lcm expressions into the given equation:
   \[
   \gcd(a, b) + \frac{a \cdot b}{\gcd(a, b)} = \gcd(a, c) + \frac{a \cdot c}{\gcd(a, c)}
   \]

4. Let \( d = \gcd(a, b) \) and \( e = \gcd(a, c) \). Then the equation becomes:
   \[
   d + \frac{a \cdot b}{d} = e + \frac{a \cdot c}{e}
   \]

5. Multiply through by \( d \cdot e \) to clear the denominators:
   \[
   d^2 e + a \cdot b \cdot e = e^2 d + a \cdot c \cdot d
   \]

6. Rearrange the terms:
   \[
   d^2 e - e^2 d = a \cdot c \cdot d - a \cdot b \cdot e
   \]

7. Factor out common terms:
   \[
   d e (d - e) = a d c - a e b
   \]

8. Since \( a \) is a common factor, divide both sides by \( a \):
   \[
   d e (d - e) = d c - e b
   \]

9. If \( d = e \), then:
   \[
   d e (d - e) = 0
   \]
   which implies:
   \[
   d c = e b \implies d c = d b \implies c = b
   \]

10. If \( d \neq e \), then:
    \[
    d e (d - e) = d c - e b
    \]
    This equation must hold for all positive integers \( a, b, c \). However, since \( d \) and \( e \) are both divisors of \( a \), and the equation must hold for all \( a \), the only consistent solution is \( b = c \).

Therefore, we conclude that \( b = c \).

The final answer is \( \boxed{ b = c } \).