Is it Possible to Find Value of a Non Symmetric Polynomial?

A symmetric polynomial  
\[f(\alpha , \beta , \gamma ,...)\]
  is a polynomimal such that if the arguments  
\[\alpha , \beta , \gamma , \...\]
  are raeearanged, then the value of  
\[f\]
  remains the same, because rearranges the arguments only changes the order of the terms.
Example:  
\[f(\alpha , \beta , \gamma )= \alpha^2 + \beta^2 + \gamma^2\]
  is a symmetric polynomial because  
\[f(\alpha , \beta , \gamma )=f(\alpha , \beta , \gamma )=f(\alpha , \beta , \gamma )= \alpha^2 + \beta^2 + \gamma^2\]
.
Suppose  
\[f(\alpha , \beta )= \alpha - \beta\]
.
 
\[f(\alpha , \beta )\]
  is not a symmetric polynomial because  
\[f(\alpha , \beta )= \alpha - \beta \neq f(\beta , \alpha )= \beta - \alpha\]
.
We can however create a symmetric polynomial from  
\[f(\alpha , \beta , \gamma )\]
  by squaring it.
\[(f(\alpha , \beta ))=(\alpha - \beta )^2 = \alpha^2 - 2 \alpha \beta \beta^2 = \beta^2 - 2 \beta \alpha + \alpha^2 = (f(\beta , \alpha ))^2 \]
.
Can we then find the value of  
\[f(\alpha , \beta ) = \alpha - \beta\]
  by finding the square root of  
\[(f(\alpha , \beta))^2\]
?
Suppose the root of the equation  
\[z^2- 3z-5\]
  are  
\[\alpha , \: \beta \]
.
Since the roots are  
\[\alpha , \: \beta\]
,  
\[(z- \alpha )(z - \beta )= z^2 - (\alpha + \beta ) z + \alpha \beta =0\]
.
Comparing with the equation above,  
\[\alpha + \beta = 3, \: \alpha \beta = -5\]
.
\[(f(\alpha , \beta ))^2 = z^2 - 2 \alpha \beta + \beta ^2 = (\alpha + \beta )^2 - 4 \alpha \beta =3^2-4 \times -5 =29\]
.
Then  
\[f(\alpha , \beta )= \pm \sqrt{29}\]
.
This will only work if the function  
\[f\]
  has terms such that powers of roots of symmetric, and the only things that stops symmetry are the signs of some terms.
The value of  
\[\alpha - 3 \beta\]
  cannot be found, and neither can the value of  
\[\alpha + 3 \beta\]

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