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 $n=2,3,5$.
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 $a_n(x)=0$.