Ta có :
$x^{6}+3x^{2}+1\leq x^{6}+3x^{4}+3x^{2}+1=(x^{2}+1)^{3}$ (vì $3x^4\geq 0$)
Mà $x^{6}+3x^{2}+1 > x^6=(x^2)^3$ (vì $3x^2+1>0$ vs mọi $x$)
=> $(x^2)^3<x^{6}+3x^{2}+1\leq (x^{2}+1)^{3}$
Mà $x^{6}+3x^{2}+1=y^{3}$
$\Rightarrow y^3=x^{6}+3x^{2}+1=(x^{2}+1)^{3}$ (do $x$ nguyên)
$\Rightarrow x=0\Rightarrow y=1$