We are to prove

if x5=x4+1, then x³=x+1

or

if x5-x4-1=0, then x³-x-1=0

or we are to prove:

if f(x) = x5 - x4 - 1 = 0  and g(x) = x³ - x - 1 = 0

then all zeros of f(x) are zeros of g(x).  

This is likely not true for all complex zeros of f(x), since a 5th degree
polynomial has 5 zeros (counting multiplicities) whereas a 3rd degree
polynomial has only 3.  However it may be true for x real.


Let's consider the case when x is a real number.

By Descartes' rule of signs, both f(x) and g(x) have 1 positive zero.  

f(x) has no negative solutions. g(x) either has 2 or no negative zeros.
They could have the same positive zero if g(x) were a factor of f(x).
So we see if this is the case by long division of f(x)÷g(x):
                                                                            
                                   x² -  x + 1
x³ + 0x² - x - 1)x5 -  x4 + 0x³ + 0x² - 0x - 1
                 x5 + 0x4 -  x³ -  x²
                      -x4 +  x³ +  x² - 0x
                      -x4 - 0x³ +  x² +  x
                             x³ + 0x² -  x - 1
                             x³ + 0x² -  x - 1
                                             0
Yes indeed we get a 0 remainder, so

x5-x4-1 = (x³-x-1)(x²-x+1)

f(x) = g(x)·(x²-x+1)

x²-x+1 has only conjugate complex imaginary zeros , so if
x = the one and only real zero of f(x), the right side is also 0, and x²-x+1 is
not 0 since it has only imaginary zeros  , so x must also
be the real zero of g(x).  So the proposition is true for x real.

However it is not true for x complex imaginary, for the zeros
 of x²-x+1 are zeros of f(x), but are not zeros of g(x).  

Edwin