Eliminate the inner mileage bracket in an empty list and do not exclude the use of cons

The goal is to remove all inner brackets.

(flatten '(a (bc) d)) becomes' (abcd)

This is my Racket code.

; if slist is null, return empty
; otherwise, if it is a pair, recursively solve car and cdr and concat them
; if it is a symbol, return the symbol

(define flatten
  (lambda (slist)
    (cond
      [ (null? slist) '()]
      [ (pair? slist)  
        (cons ((flatten (car slist)) (flatten (cdr slist))))]
      [ (symbol? slist) slist])))

He complains

procedure application: expected procedure, given: c; arguments were: ()

which means that I'm trying to access carand cdrfor an empty list.

I did the trace:
> (flatten '(a (b c) d))
pair?-car-cdr
a
((b c) d)
symbol?
a
pair?-car-cdr
(b c)
(d)
pair?-car-cdr
b
(c)
symbol?
b
pair?-car-cdr
c
()
symbol?
c
(stops here)

The trace code is simple - a bunch of displays.

(define flatten
  (lambda (slist)
    (cond
      [ (null? slist) '()]
      [ (pair? slist) 
        (display 'pair?-car-cdr)
        (newline)
        (display (car slist))
        (newline)
        (display (cdr slist))
        (newline)
        (cons ((flatten (car slist)) (flatten (cdr slist))))]
      [ (symbol? slist) 
         (display 'symbol?)
         (newline)
         (display slist)
         (newline)
        slist])))

I do not understand why the first condition (null? slist)did not catch an empty list? I have two recursive calls. If he caught an empty list, he will move on to the next recursion, which is a list {d}.

What is the problem with my recursion logic?

Update version

(define flatten
  (lambda (slist)
    (cond
      [ (null? slist) '()]
      [ (pair? slist)  
        (cons (flatten (car slist)) (flatten (cdr slist)))]
      [ (symbol? slist) slist])))

(display (equal? (flatten '(a (b a) b a c (a b) c (e f (b a)))) '(a b a b a c a b c e f b a)))
(newline)
(display (equal? (flatten '(a b c)) '(a b c)))
(newline)
(display (equal? (flatten '(a (b c))) '(a b c)))
(newline)
(display (equal? (flatten '((a)(b)(c) d)) '(a b c d)))
(newline)
(display (equal? (flatten '(a (b) ((c)) (((d))) ((((e (f g))))))) '(a b c d e f g )))
(newline)
(display (equal? (flatten '()) '()))
(newline)
(display (equal? (flatten '(a b () ())) '(a b)))
(newline)

, , . , , , (2- )

-, (cons (flatten slist) empty)