<s>[INST] Another simple rule satisfied by the Fibonacci $n$-step constants? Given,
$$x^n(2-x)=1\tag1$$
for $n=2,3,4,\dots$
$$(x-1)(x^2-x-1)=0\\
(x-1)(x^3-x^2-x-1)=0\\
(x-1)(x^4-x^3-x^2-x-1)=0$$
the roots of which are the golden ratio, the tribonacci constant, the tetranacci constant, and so on. 

Q: Is it true that for all integer $n>1$, if $y=x^{-1}$ then,

$$2y\,(1-y^{n-1})(1-y^{2n+2}) = (1-y^{n+1})^3\tag2$$
and
$$(1-y^{n-1})(1-y^{2n+2})^2 = (1-y^{n+1})^4\tag3$$
such that the RHS is a cube and fourth power, respectively?
P.S. See also this post for special relations for $n=2,3,5$.
 [/INST] For all integer $n$ define $a_n(x) := 1-2x^n+x^{n+1} = (x-1)(x^n - \dots -x^2 - x - 1).$
For $y=x^{-1}$ we have the two algebraic identities
$$ \frac{a_n(x)(1-x^{n+1})(1+2x+x^{n+1})}{x^{3n+3}} = 
2y(1-y^{n-1})(1-y^{2n+2}) - (1-y^{n+1})^3,$$
$$ \frac{-a_n(x)(1 - x^{n + 1})^2(1 + 2 x^n + x^{n + 1})}{x^{5n+3}} =
(1-y^{n-1})(1-y^{2n+2})^2 - (1-y^{n+1})^4$$
which prove equations (2) and (3), respectively, if we assume $a_n(x)=0$.
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