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Is this enough for a proof?:

$$x^3+x^2 = 1$$

I would factor and get: $x^2(x+1) = 1$

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I would show that $x = \sqrt1$, which is irrational but then do I have to show more? $x+1=1$ which gives me $x=0$ and since $x$ cannot equal to $0$ as this would make the statement false (everything is $0$). Is it enough to simply state this falsity or is there another way to express it?

Thanks!

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By the rational root theorem, a rational root would have to be $x=1$ or $x=-1$, but neither works.

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}

$$

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 $A$).

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.

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