y³ - 9x² + 27x - 27 = 0
z³ - 9y² + 27y - 27 = 0
x³ - 9z² + 27z - 27 = 0

Since all the equations will be the same if x = y = z,
that must give a solution:

x³ - 9x² + 27x - 27 = 0

Possible zeros are ±1, ±3, ±9, ±27

We find that when we try x = 3

3 | 1  -9  27  -27
  |     3 -18   27
    1  -6   9    0

we can factor it as

(x - 3)(x² - 6x + 9) = 0

and factor it further as

(x - 3)(x - 3)(x - 3) = 0

             (x - 3)³ = 0

                x - 3 = 0
    
                    x = 3

Thus x=3, y=3, z=3 is a solution.  

There are 5 complex solutions as well:

x=-9.97528-0.398972i, y=-5.51359+9.06114i, z=0.872034+10.5311i
x=-9.97528+0.398972i, y=-5.51359-9.06114i, z=0.872034-10.5311i
x=-8.0106-5.47326i, y=9.67166+3.49043i, z-6.35574+6.29188i
x=-8.0106+5.47326i, y=9.67166-3.49043i, z=-6.35574-6.29188i
x=-6.35574-6.29188i, y=-8.0106+5.47326i, z=9.67166-3.49043i

but I think finding those is beyond the scope of your course.
 
Edwin