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How to solve $$ \begin{cases} abc=xyz\\ a+b+c=x+y+z\\ ab+bc+ac=xy+yz+xz\\ \end{cases} $$ ?

We have $$ ab+bc+ac=xy+yz+xz\implies abc+bc^2+ac^2=c(xy+yz+xz) $$ and $$ a+b+c=x+y+z\implies ac^2+bc^2+c^3=c^2(xy+yz+xz) $$ So $$ xyz-c^3 = c(xy+yz+xz)-c^2(xy+yz+xz). $$ But this is too difficult, could you please give me a better solution? Thanks.

Ryze
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    Suppose that there are two cubic monic polynomials $f(X)$ and $g(X)$ each having the roots ${a,b,c}$ and ${x,y,z}$ respectively. By Vieta's formulas we know that the coefficients of the two polynomials are equal and hence the roots are the same. So the only solutions come from $${a,b,c}={x,y,z}$$ – Peter Foreman Jun 20 '20 at 12:43
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    @PeterForeman Many thanks! – Ryze Jun 20 '20 at 12:45

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