Proof By Contradiction, Rational Roots This was an exam question that I got totally wrong and am a bit question. 
Prove $x^3 + x + 1 = 0$ has no solutions. Prove by contradiction. 
Assume: $x^3 +x +1 =0$ has at least one rational root.
So, what I attempted to do here was solve for $x$. 
$\begin{align} x^3 +x &= -1\\
x(x^2 +1) &= -1\\
x= -1 &\vee x = \sqrt {-2}\end{align}$ 
I noticed in other answers I found online that the way to do this is to get to $a^3 + ab^2 + b^3 = 0$ and then go through each scenario for $a$ and $b$ (odd or even.) And this makes much more sense in retrospect. But can someone explain why solving for $x$ is not an appropriate starting point?     
 A: As an odd-degree polynomial, $x^3+x+1=0$ certainly has real solutions, but none that are rational. Applying the rational root theorem shows us that the only possible rational roots are $\pm1$, and substitution shows us that neither of these works.
It's worth noting that you can't conclude from $a\cdot b=-1$ that $a=-1$ or $b=-1$. For example, let $a=2,b=-\frac12$. The only time that we can conclude (for real numbers $a,b,c$) that $ab=c$ implies $a=c$ or $b=c$ is when $c=0$.
In particular, the only real solution to $x(x^2+1)=-1$ (or equivalently, $x^3+x+1=0$) is $$x=\sqrt[3]{\frac{\sqrt{93}-9}{18}}-\sqrt[3]{\frac{\sqrt{93}+9}{18}},$$ which isn't remotely obvious by your method. Who would even guess that? We can find this through Cardano's method.
A: Just because $a\,b=-1$, you can't conclude that either $a$ or $b$ must equal $-1$.
A: No solution or no rational solution? All odd degree polynomial have at least one real root. Solving for $x$ would be fine, but your solution is wrong. From $x(x^2+1)=-1$ you can't conclude $x=-1$ or $x=\sqrt{-2}$
A: A few things to point out. First, if you know the rational root theorem, then it is easy to see that $\pm 1$ are the only possible candidates for rational roots. 
That being said, it seems you are very confused by the procedure we usually use to separate out the roots of a polynomial.
Normally we take a polynomial and attempt to factor it. For example
$$0=x^2 + 7x + 12 = (x+3)(x+4)$$
Now we are able to conclude that either 
$$(x+3)=0\ \ \text{or}\ \ \ (x+4)=0$$
This is not an arbitrary step, but rather it comes from the fact that when you're working with real numbers, if $ab=0$ then either $a=0$ or $b=0$. More formally, $\mathbb{R}$ has no zero divisors because it is a field.
To me, it looks like what you've tried to do is to pull the same trick where the right hand side is not zero, i.e. you tried to conclude something analagous to
$$(x+3)(x+4)=1 \implies (x+3)=1\ \ \text{or}\ \ \ (x+4)=1$$
That is simply untrue. There is nothing which says that if $ab=c$ for $c\neq 0$ that $a=c$ or $b=c$.
A: The problem is that $ab = -1$ doesn't imply "$a = -1$ or "b = -1". If you knew $a,b$ were integers, you could infer that "$a=1$ and $b=-1$ or $a=-1$ and $b=1$" because $a$ must be a divisor of $-1$ and there are only two possibilities, but that's not the case here.
Your initial idea can be salvaged though; if you suppose $a/b$ is a rational root with $\gcd(a,b) = 1$, then your approach to the problem is to rewrite it as
$$ a (a^2 + b^2) = -b^3 $$
Now, both sides are integers and we can do something useful; we know that $a$ must be a divisor of $-b^3$; given our hypothesis there are only two possibilities for $a$. You can continue to make similar divisibility arguments until you have excluded all possibilities.
