Conflicting situations arising while solving for $x$

Question

Given $$x^{1/4}+\frac{1}{x^{1/4}}=1$$

find $$x^{1024}+\frac{1}{x^{1024}}=?$$

My approach let $a=x^{1/4}$ $$a+\frac{1}{a}=1$$ $$a^2+1-a=0 $$ multiplying by $(a+1)$ $$(a+1)(a^2+1-a)=0 $$ $$a^3+1=0, a^3=-1$$ by solving we get value of $x$ $$x^3=1$$ Now the answer to the should be 2. $$x^{1024}+\frac{1}{x^{1024}}=2$$

But if we square the equation $$x^{1/4}+\frac{1}{x^{1/4}}=1$$ we will get $-1$ as the recurring digit on the RHS as we square on both the sides. Thus it will result in $-1$ as the answer. Can anyone help me to figure out where I am wrong

Answer 1

Since $a^3=-1$, one has $a^6=1$ and $$ x=(x^\frac14)^4=a^4=-a. $$ So $$ x^{1024}+\frac{1}{x^{1024}}=a^{1024}+\frac{1}{a^{1024}}=a^{6\cdot17+4}+\frac{1}{a^{6\cdot17+4}}=a^4+\frac{1}{a^4}=-a-\frac{1}{a}=-1$$

Answer 2

$\sqrt[4]x$ and $\frac{1}{\sqrt[4]x}$ are roots of the equation $$z^2-z+1=0,$$ they are $\cos60^{\circ}+i\sin60^{\circ}$ and $\cos60^{\circ}-i\sin60^{\circ}$.

Id est, $$x^{1024}+\frac{1}{x^{1024}}= (\cos60^{\circ}+i\sin60^{\circ})^{4096}+(\cos60^{\circ}-i\sin60^{\circ})^{4096}=$$ $$=(\cos60^{\circ}+i\sin60^{\circ})^{4+682\cdot6}+(\cos60^{\circ}-i\sin60^{\circ})^{4+682\cdot6}=$$ $$=\cos240^{\circ}+i\sin240^{\circ}+\cos240^{\circ}-i\sin240^{\circ}=-1.$$