Prove that $x^3 + x^2 = 1$ has no rational solutions? Is this enough for a proof?:
$$x^3+x^2 = 1$$
I would factor and get: $x^2(x+1) = 1$
I would show that $x = \sqrt1$, which is rational but then what else would I have to show? $x+1=1$ which gives me $x=0$ and since $x$ cannot equal to $0$ as this would make the statement false ($0$ times anything is $0$). Is it enough to simply state this falsity or is there another way to express it?
Thanks!
 A: By the rational root theorem, a rational root would have to be $x=1$ or $x=-1$, but neither works.
A: Let's assume $x = p/q$. $p$ and $q$ integers without a common factor. Then,
$$
p^{3} + p^{2}q = q^{3}
$$
It's is only satisfied whenever $p$ and $q$ are simultaneously even. It contradicts the initial hypothesis that we can set $x = p/q$ where $p$ and $q$ has not common factors.
$$
\mbox{Then,}\quad x \not\in {\mathbb Q}
$$
A: Theorem
The solution satisfying the following equation
$$
A \times B =0
$$
is $A=0$ (for any $B$) or $B=0$ (for any $A$).
You cannot apply the same pattern for the case in which the right hand side is not zero. Why?
For example,
$$
A\times B = 2
$$
If you choose $A=2$ then $B$ must be $1$ (rather than for any $B$). If you choose $B=2$ then $A$ must be $1$ (rather than for any $A$). 
Back to your question
If you want to find the solution of
$$x^2(x+1) =1$$
you have to make sure the right hand side equals to 0.
\begin{align*}
x^2(x+1) &=1\\
x^3 + x^2 -1 &=0
\end{align*}
To prove the equation has no rational solution see this comment.
