If $(1-y)\left(1+2x+4x^2+8x^3+16x^4+32x^5)=1-y^6,(y\ne 1)\right.$, then a value $y/x$ is

We can rewrite the given expression as
$(1-y)\frac{\left[1-(2x{)}^6\right]}{1-2x}=1-y^6,$
one of the possible values of $y$ is clearly $2x$. Therefore, one of the possible values of $y/x$ is $2.$