Solutions in a field Suppose $F$ is a field. Suppose further that $F$ is an ordered field. Consider the following equation $x^3 = b$, where $b \in F$. Prove that the equation above has, at most, one solution.
My attempt:
Suppose we have $x = b^{1/3} \in F$. We have that
$$x^3 = (b^{1/3})^3) = b^{\frac{1}{3} \cdot 3} = b^1 = b$$
Thus, there is one solution.
Assume there exists another solution and denote this by $c^{1/3}$. Since $b^{1/3}$ is a solution, $x = b^{1/3} \implies (b^{1/3})^3 = b \implies (b^{1/3})^3 - b = 0$. In a similar fashion, since $c^{1/3}$ is another solution, $x = c^{1/3} \implies (c^{1/3})^3 = b \implies (c^{1/3})^3 - b = 0$. Together we see that
$$(b^{1/3})^3 - b = 0 = (c^{1/3})^3 - b \implies (b^{1/3})^3 =(c^{1/3})^3 \implies b^{1/3} = c^{1/3} \implies b = c$$
Therefore there is at most one solution.
Is this correct for a field, $F$? Or did I make a mistake? Any assistance is much appreciated.

You offer no justification for your claim that $b^{1/3} = c^{1/3} \implies b = c$.

In essence, that's what the problem requires you to prove.

As an indication that your proof is not valid, note that you never used the hypothesis that $F$ is an ordered field.

Thus, let $F$ be an ordered field, and suppose $x,y\in F$ are such that $x^3=y^3$.

We want to show $x=y$.

If $x=0$, then $x^3=0$, so $y^3=0$, hence $y=0$, so $x=y$.

Similarly, if $y=0$, then $y^3=0$, so $x^3=0$, hence $x=0$, so $x=y$.

Thus suppose $x,y\ne 0$, and assume $x\ne y$.

Our goal is to derive a contradiction.

\begin{align*}
\text{Then}\;\;&
x^3=y^3
\\[4pt]
\implies\;&
x^3-y^3=0
\\[4pt]\implies\;&
(x-y)(x^2+xy+y^2)=0
\\[4pt]
\implies\;&
x^2+xy+y^2=0
\\[4pt]
\implies\;&
4x^2+4xy+4y^2=0
\\[4pt]
\implies\;&
(4x^2+4xy+y^2)+3y^2=0
\\[4pt]
\implies\;&
(2x+y)^2+3y^2=0
\\[4pt]
\end{align*}
conradiction, since $(2x+y)^2 \ge 0$ and $3y^2 > 0$.
