How to avoid overflow in expr. A * B - C * D

This should work ( I think ):

signed long long int a = 0x7ffffffffffffffd;
signed long long int b = 0x7ffffffffffffffd;
signed long long int c = 0x7ffffffffffffffc;
signed long long int d = 0x7ffffffffffffffe;
signed long long int bd = b / d;
signed long long int bdmod = b % d;
signed long long int ca = c / a;
signed long long int camod = c % a;
signed long long int x = (bd - ca) * a * d - (camod * d - bdmod * a);

Here's my derivation:

x = a * b - c * d
x / (a * d) = (a * b - c * d) / (a * d)
x / (a * d) = b / d - c / a

now, the integer/mod stuff:
x / (a * d) = (b / d + ( b % d ) / d) - (c / a + ( c % a ) / a )
x / (a * d) = (b / d - c / a) - ( ( c % a ) / a - ( b % d ) / d)
x = (b / d - c / a) * a * d - ( ( c % a ) * d - ( b % d ) * a)

Note that this is not standard since it relies on wrap-around signed-overflow. (GCC has compiler flags which enable this.)

But if you just do all the calculations in long long, the result of applying the formula directly:
(A * B - C * D) will be accurate as long as the correct result fits into a long long.

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Here's a work-around that only relies on implementation-defined behavior of casting unsigned integer to signed integer. But this can be expected to work on almost every system today.

(long long)((unsigned long long)A * B - (unsigned long long)C * D)

This casts the inputs to unsigned long long where the overflow behavior is guaranteed to be wrap-around by the standard. Casting back to a signed integer at the end is the implementation-defined part, but will work on nearly all environments today.

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If you need more pedantic solution, I think you have to use "long arithmetic"

The simplest and most general solution is to use a representation that can't overflow, either by using a long integer library (e.g. http://gmplib.org/) or representing using a struct or array and implementing a kind of long multiplication (i.e. separating each number to two 32bit halves and performing the multiplication as below:

(R1 + R2 * 2^32 + R3 * 2^64 + R4 * 2^96) = R = A*B = (A1 + A2 * 2^32) * (B1 + B2 * 2^32) 
R1 = (A1*B1) % 2^32
R2 = ((A1*B1) / 2^32 + (A1*B2) % 2^32 + (A2*B1) % 2^32) % 2^32
R3 = (((A1*B1) / 2^32 + (A1*B2) % 2^32 + (A2*B1) % 2^32) / 2^32 + (A1*B2) / 2^32 + (A2*B1) / 2^32 + (A2*B2) % 2^32) %2^32
R4 = ((((A1*B1) / 2^32 + (A1*B2) % 2^32 + (A2*B1) % 2^32) / 2^32 + (A1*B2) / 2^32 + (A2*B1) / 2^32 + (A2*B2) % 2^32) / 2^32) + (A2*B2) / 2^32

Assuming the end result fits in 64 bits you actually don't really need most bits of R3 and none of R4

This seems too trivial I guess. But A*B is the one that could overflow.

You could do the following, without losing precision

A*B - C*D = A(D+E) - (A+F)D
          = AD + AE - AD - DF
          = AE - DF
             ^smaller quantities E & F

E = B - D (hence, far smaller than B)
F = C - A (hence, far smaller than C)

This decomposition can be done further.
As @Gian pointed out, care might need to be taken during the subtraction operation if the type is unsigned long long.

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For example, with the case you have in the question, it takes just one iteration,

 MAX * MAX - (MAX - 1) * (MAX + 1)
  A     B       C           D

E = B - D = -1
F = C - A = -1

AE - DF = {MAX * -1} - {(MAX + 1) * -1} = -MAX + MAX + 1 = 1