Formula:

$$ (f,d,b) = (d(f-1) \mod{(b-1)}+1)\times\left\lfloor\frac{b^{\left\lfloor\dfrac{d(f-1)}{b-1}\right\rfloor+1}}{b-(b>f)(b>2)}\right\rfloor $$

<                   # Decrease the first (implicit) input `f` by 1
 *                  # Multiply it to the second (implicit) input `d`
    ‰               # Divmod it by:
  I<                # The third input `b` minus 1
     >              # Increase both values in the pair by 1 (let's call them [A,B])
      `             # Pop and push them separated to the stack
       Š            # Tripleswap the stack from {A,B} to {B,(implicit) third input `b`,A}
        m           # Take `b` to the power `A`
         I          # Push the third input `b` again
          D         # And again
           ¹2‚      # Push pair [f,2]
              ›     # Pop the copy of `b`, and check [b>f, b>2]
               P    # Check if both are truthy by taking the product
                -   # Decrease the other `b` by this 0 or 1
                 ÷  # Integer-divide `b to the power A` by this
                  * # Multiply it to `B`
                    # (after which the result is output implicitly)

Formula: (d*(f-1)%(b-1)+1)*((b**(d*(f-1)//(b-1)+1))//(b-(b>f)*(b>2)))

<                   # Decrease the first (implicit) input `f` by 1
                    #  STACK: f-1
 *                  # Multiply it to the second (implicit) input `d`
                    #  STACK: d*(f-1)
    ‰               # Divmod it by
  I<                # the third input `b` minus 1
                    #  STACK: [d*(f-1)//(b-1), d*(f-1)%(b-1)]
     >              # Increase both values in the pair by 1 (let's call them [A,B])
                    #  STACK: [d*(f-1)//(b-1)+1, d*(f-1)%(b-1)+1]
      `             # Pop and push them separated to the stack
                    #  STACK: d*(f-1)//(b-1)+1, d*(f-1)%(b-1)+1
       Š            # Tripleswap the stack using the implicit third input `b`
                    #  STACK: d*(f-1)%(b-1)+1, b, d*(f-1)//(b-1)+1
        m           # Take `b` to the power `A`
                    #  STACK: d*(f-1)%(b-1)+1, b**(d*(f-1)//(b-1)+1)
         I          # Push the third input `b` again
                    #  STACK: d*(f-1)%(b-1)+1, b**(d*(f-1)//(b-1)+1), b
          D         # And again
                    #  STACK: d*(f-1)%(b-1)+1, b**(d*(f-1)//(b-1)+1), b, b
           ¹2‚      # Push pair [f,2]
                    #  STACK: d*(f-1)%(b-1)+1, b**(d*(f-1)//(b-1)+1), b, b, [f,2]
              ›     # Pop the copy of `b`, and check [b>f, b>2]
                    #  STACK: d*(f-1)%(b-1)+1, b**(d*(f-1)//(b-1)+1), b, [b>f,b>2]
               P    # Check if both are truthy by taking the product
                    #  STACK: d*(f-1)%(b-1)+1, b**(d*(f-1)//(b-1)+1), b, (b>f)*(b>2)
                -   # Decrease the other `b` by this 0 or 1
                    #  STACK: d*(f-1)%(b-1)+1, b**(d*(f-1)//(b-1)+1), b-(b>f)*(b>2)
                 ÷  # Integer-divide `b to the power A` by this
                    #  STACK: d*(f-1)%(b-1)+1, (b**(d*(f-1)//(b-1)+1))//(b-(b>f)*(b>2))
                  * # Multiply it to `B`
                    #  STACK: (d*(f-1)%(b-1)+1)*((b**(d*(f-1)//(b-1)+1))//(b-(b>f)*(b>2)))
                    # (after which the result is output implicitly)