linear equations

Question

1) Find 3 nos. If the cube of the first number exceeds theur product by 2, the cube of the second no. is smaller than their product by 3 and the cube of the third no. exceeds their product by 3

Answer 1

Adding the second and third:

y³ + z³ = 2xyz

Adding the first and second

x³ + y³ = 2xyz - 1

Substituting from above

x³ + y³ = y³ + z³ - 1

x³ = z³ - 1

 

Know any numbers whose cubes differ by one?

Answer 2

I believe that you mean (assuming three numbers are x, y and z)...

x³ = xyz + 2

y³ = xyz - 3 and 

z³ = xyz + 3.

 

This looks like a rather hard system to solve. Have you tried TRIAL & ERROR?  (I suggest that you could try an Excel spreadsheet in which different small integers were chosen for (x,y,z) triples, and then have various columns computing their cubes and the other three right hand sides above.)