Prove that there is no rational root of $x^3+x+1=0$ by contradiction

Question

I have found that this equation only has one root between $-1$ and $0$. I try to let this root=$p/q$ where $gcd(p,q)=1$ and get $p^3+pq^2+q^3=0$. I want to know how to prove it by the method of contradiction.

Answer 1

You did a great job. Only a couple of steps left.

$0=p^3+pq^2+q^3=p(p^2+q^2)+q^3 \implies p|q^3$, but $gcd(p,q)=1$, which means that $p=\pm 1$. Similarly $0=p^3+pq^2+q^3=q(pq+q^2)+p^3 \implies q|p^3$, but $gcd(p,q)=1$, which means that $q=\pm 1$, therefore $x=\pm 1$. But $f(1)=3 \neq 0$ and $f(-1)=-1 \neq 0$, so there is no rational root.