How can I prove that if $a^3=b^3\rightarrow a=b$ using only field axioms $\forall a,b\in\mathbb{R}$.
I feel like it has something to do with $$ a^3-b^3=0\rightarrow a^3+(-a^3)=0 \rightarrow -b^3=(-a^3) $$ but I have no clue how to continue the proof.
I am also not sure if it's possible to say that $$ a^3-b^3=(a - b)\cdot(a^2 + ab + b^2) $$ because I am restricted to prove this with field axioms only.
HINT
According to Newton's binomial formula, one has: \begin{align*} (a - b)^{3} & = a^{3} - 3a^{2}b + 3ab^{2} - b^{3} = (a^{3} - b^{3}) - 3ab(a - b) \end{align*} whence we conclude that \begin{align*} a^{3} - b^{3} & = (a - b)^{3} + 3ab(a - b)\\\\ & = (a - b)[(a - b)^{2} + 3ab]\\\\ & = (a - b)(a^{2} + ab + b^{2}) \end{align*}
As @JoséCarlosSantos has mentioned, it is necessary to consider the fact that $\mathbb{R}$ is linearly ordered (which means that it is a partially ordered set and $x\in\mathbb{R}$ and $y\in\mathbb{R}$ are always comparable): \begin{align*} a^{2} + ab + b^{2} & = \left(a^{2} + ab + \frac{b^{2}}{4}\right) + \frac{3b^{2}}{4}\\\\ & = \left(a + \frac{b}{2}\right)^{2} + \frac{3b^{2}}{4} \geq 0 \end{align*}
Can you take it from here?