若y=0 则x=1 显然可以验证成立
若y0 则由于x^3-2y^3=1 所以(x^3)/(y^3)-2=1/(y^3)
即(x/y)^3-2=1/(y^3) 从而 (x/y)^3-2=[x/y-2^(1/3)]*[(x/y)^2+(x/y)*2^(1/3)+4^(1/3)]=1/(y^3)
从而|(x/y)-2^(1/3)| * | [(x/y)^2+(x/y)*2^(1/3)+4^(1/3) |=1/|y^3|;
令x/y=t t^2+t*2^(1/3)+4^(1/3)>=(3/4)*4^(1/3) 从而| [(x/y)^2+(x/y)*2^(1/3)+4^(1/3) |>1/4
从而|(x/y)-2^(1/3)|