Roots of polynomial equation $x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$

Question

If $x_1,x_2,...,x_6$ be the roots of $x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$ then I have to show that $|x_j|=2\space\space\space\forall j\in\{1,2,3,4,5,6\}$

I get that the roots of the equation must be complex, of the form $a+ib$ where $|a+ib|=\sqrt{a^2+b^2}=2$ for all $|x_i|$. I don't get how to find the value of $a+ib$. How do I find the roots?

Answer 1

Hint:\begin{multline}x^6+2x^5+4x^4+8x^3+16x^2+32x+64=\\=64\left(\left(\frac x2\right)^6+\left(\frac x2\right)^5+\left(\frac x2\right)^4+\left(\frac x2\right)^3+\left(\frac x2\right)^2+\frac x2+1\right).\end{multline}Also, use the fact that $x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)$.

Answer 2

$$x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$$ Let $z=x/2$, so $$z^6+z^5+z^4+z^3+z^2+z+1=0$$ from $z \neq 1$ Then $\frac{1-z^7}{1-z}=0 $ so $1-z^7=0$ so $z$ is the $n$ th root of $7$. from $x=2z$ ,so $|x|=|2z|$ so $|x|=|2z|=2$