True:

Proof:

Case 1:
If a > b and a and b are both positive, then
there exists positive x such that

b = a + x

Cubing both sides of the equation 

a + x = b

(a+x)^3 = b^3

a^3 + 3a^2x + 3ax^2 + x^3 = b^3

Then k = 3a^2x + 3ax^2 + x^3 is positive, therefore 

there exists positive number k so that a^3 + k = b^3

Case 2:
If a > b and a is negative and b is positive, then a^3 
is negative and b^3 is positive and any positive number 
is greater than any negative number.

Case 3:
If a > b and a is negative and b is negative, then -b > -a > 0, and by
Case 1 

(-b)^3 > (-a)^3
-b^3 > -a^3
a^3 > b^3

Case 4:

If a > b and b = 0  Then a > 0 and  a^3 > 0, so b^3 > a^3

Case 5:

If a > b and a = 0, then  b < 0 and b^3 < 0 so b^3 < a^3 or a^3 > b^3   


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False.  Here is a counter-example to disprove it:

a = 2, b = 1, so a > b

c = 5, d = 3, so c > d

c - a = 5 - 2 = 3
d - b = 3 - 1 = 2

However  c - a > d - b since 3 > 2

Edwin