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Let $a,b,c,x,y,z$ be any real numbers. I want to show that $\big\{\{a\},\{b\},\{c\}\big\}=\big\{\{x\},\{y\},\{z\}\big\}$ if and only if \begin{align*} x+y+z & =a+b+c \\ xy+xz+yz & =ab+ac+bc \\ xyz & =abc \end{align*} Clearly if $\big\{\{a\},\{b\},\{c\}\big\}=\big\{\{x\},\{y\},\{z\}\big\}$ the previous equalities hold. How could I prove the converse ? Any help or hint is very much appreciated.

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    Remember Vieta's relations. – 2017-02-01
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    It's only true if $a,b,c$ are distinct. For example, $\{1,1,2\}=\{1,2,2\}$ but the formulas are not true. – 2017-02-01
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    @Thomas Andrews: is the first equation true for that example? – 2017-02-01
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    @ThomasAndrews Right! Perhaps asking instead for $\{\{a\},\{b\}\{c\} \}=\{\{x\},\{y\}\{z\}\}$ fixes the issue without asking $a,b,c$ to be distinct. – 2017-02-01
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    What is the first question? $\{1,1,2\}=\{1,2,2\}$ under the usual meaning of those symbols. @CarlMummert – 2017-02-01
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    @LeoSera Don't think that $a \mapsto \{a\}$ changes much there. The more common way to phrase it would be "*the triplets (or tuples) are identical up to permutation*". – 2017-02-01
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    @Leo Sera: $\{\{a\},\{b\},\{c\}\} = \{\{x\},\{y\},\{z\}\}$ if and only if $\{a,b,c\} = \{x,y,z\}$. For example, assuming the first equation, we have $\{a\}$ as a member of $\{\{x\},\{y\},\{z\}\}$, so $a \in \{x,y,z\}$, and conversely if $a \in \{x,y,z\}$ then $\{a\} \in \{\{x\}, \{y\}, \{z\}\}$. – 2017-02-01
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    @Thomas Andrews: I overlooked the word "not" in your comment - apologies. That does give a correct counterexample, of course, with $a = 1$, $b =1$, $c = 2$, $x = 1$, $y = 2$, $z = 2$. I had been looking for a counterexample at the time, so I was interested in yours. – 2017-02-01
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    @Leo Sera: as Thomas Andrews points out, the sentence beginning "clearly" isn't right. – 2017-02-01

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I'll take the liberty of rephrasing the question according to what it probably means, since it seems to me that the OP is missing the technical tools to phrase it correctly:

Let $(x,y,z)$ and $(a,b,c)$ be two triplets of complex numbers. Prove that $$(a,b,c)\in\{(x,y,z),\,(x,z,y),\,(y,x,z),\,(y,z,x),\,(z,x,y),\,(z,y,x)\}$$ if and only if $$\begin{cases} x+y+z=a+b+c\\ xy+xz+yz=ab+ac+bc\\ xyz=abc\end{cases}$$

The non-obvious implication is: consider the polynomials $p(T)=(T-a)(T-b)(T-c)$ and $q(T)=(T-x)(T-y)(T-z)$. If you run the calcs, you'll notice that your condition is exactly $p=q$. According to your knowledge of algebra, you should know that the roots of a polynomial are well-defined.

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    I knew that the condition seemed familiar. Thanks for the hint! – 2017-02-01
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    This doesn't answer the question, though, because just because two polynomials have the same *set* of roots does not mean that they are the same polynomial; the multiplicities would also need to be the same. – 2017-02-01