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
