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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.

LL 3.14
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jc426
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    You can't do that using only field axioms since there are fields (such as $\Bbb C$) for which that assertion is false. But it is true for every ordered field. – José Carlos Santos Oct 22 '22 at 22:10
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    Your second equation is easy to prove though, just expand the product ... – LL 3.14 Oct 22 '22 at 22:13
  • @LL3.14 I realize that but to be able to use the second equation I would have to prove that $a^3-b^3=(a-b)\cdot(a^2+ab+b^2)$ which would probably be the easiest way to go right? – jc426 Oct 22 '22 at 22:22

1 Answers1

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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?

Átila Correia
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    Using Newton's binomial formula seems a bit overkill since then one could also directly use the identity for $a^3-b^3$ ... the formula for $a^3-b^3$ can be obtained very easily from fields axiom by expanding $(a-b)(a^2+ab+b^2)$. For the other part, I agree :) – LL 3.14 Oct 22 '22 at 22:51
  • @LL3.14 Thanks for the feedback :) – Átila Correia Oct 24 '22 at 22:22