Prove $x^3 + x +1 = 0$ has no rational solutions. 
Use a suitable method to prove that there is no rational solutions to the equation:
$$x^3 + x + 1 = 0$$ or give a counter-example to show otherwise.

I tried to do this question via polynomial factorization and contradiction.
Let $$f(x) = x^3 + x + 1 = 0$$
By polynomial factorisation by assuming there is a root $r$,
$$f(x) = (x-r)(x^2 + xr + r^2 + 1)$$
$$f(x) = (x^3 + rx^2 + xr^2 + x - rx^2 - xr^2 - r^3 - r)$$
$$f(x) = (x^3 + x - r^3 - r)$$
Comparing the coefficients:
$$r^3 - r = 1$$
From this, we can assume that root $r$ exists, and hence, we have proven by contradiction that a root exists.
I want to ask if my proof is correct.
 A: (I do think the question is interesting: not because it's particularly insightful, but because it has a flaw commonly committed by newcomers)
Your proof is not correct, for two reasons:

*

*A minor computational error, that gives the illusion that you found out something.

*A major logical error.

Let's start with the minor one:
From $f(x) = x^3+x+1=(x-r)(x^2 + xr + r^2 + 1)=x^3 + x - r^3 - r$, you write that, by comparing coefficients, you get $r^3 - r = 1$. It's wrong.
Actually, by comparing coefficients, you have
$$x^3+x+1=x^3+x-r^3-r$$
$$1=-r^3-r$$
$$r^3+r+1=0$$

Now, the major problem. You would like to find a proof by contradiction.
Initially, you want to prove that the polynomial $x^3+x+1$ has no rational root. That is, any complex number $r$ such that $r^3+r+1=0$ satisfies $r\notin \Bbb Q$.
To find a proof by contradiction, you would have to assume $r$ is a root, and $r$ is rational, and find a contradiction.
Here, you only assumed that $r$ is a root. There is no contradiction to be found, as any polynomial with complex coefficients and degree $n>0$ has exactly $n$ complex roots. It has roots, hence there will be no contradiction assuming it has a root.
With your assumption, and after correcting the little computational mistake, you find out that $r^3+r+1=0$. But that's the assumption! You assumed $r$ is a root of $x^3+x+1$, i.e. exactly that $r^3+r+1=0$, and you find out that the same equality is true.

So your proof by contradiction in this current state is moot. It's still possible to do it though, using the idea behind the more general rational root theorem.
Assume $r$ is a rational root of $x^3+x+1$. That is, there are coprime integers $a$ and $b$ such that $r=a/b$ and $r^3+r+1=0$. The important fact, here, is that a rational number is the quotient of two coprime integers.
Then:
$$\frac{a^3}{b^3}+\frac ab+1=0$$
$$a^3+ab^2+b^3=0$$
Now, we have easily a contradiction: since the sum is zero and two of the three terms are divisible by $b$, then the third must also be divisible by $b$. However, per our assumtpions, it's coprime to $b$.
