p3- q3= r! -18 where p?q and r are nonnegative integers

Question

Is this even possible?

 

Answer 1

p^3-q^3= r!-18

r,q,p all >0 and are all integers

1!=1, 1-18=-17

2!=2, 2-18= -16

3! = 6, 6-18 = -12

4! = 24, 24-18 = 6

5! = 120, 120-18 = 102

6!= 720, 720-18= 702

6^3=108

7^3= 343

8^3= 514

9^3= 729

9^3-3^3=729-27=702

=6!-18=720-18 =702

p=9, q=3, r=6

either p and q are both odd, or both are even