## 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
$\alpha + 3 \beta$