Lets say I have a matrix A with constants x and paramater Y in the following form:
$$A= \begin{bmatrix} Y & x_{12} & x_{13} \\ x_{12} & Y & x_{23} \\ x_{13} & x_{23} & Y \end{bmatrix} $$
I know that if all x are the same, then $\det[A] = 0$ has only one solution for Y.
My proof here is to directly compute the determinant, and then solve the suppressed cubic.
I want to also prove that if not all x are the same, that det[A]=0 has 3 distinct solutions. This is slightly more than just the converse. Directly computing is very painful, and I was able to show it only with Mathematica.
My question is how to justify these two statements more elegantly.
Your first proposition is generally incorrect. If $x_{12}=x_{13}=x_{23}=x$, then both $Y_{1,2}=x$ and $Y_3=-2x$ are solutions of $\det[A]=0$. Those are different for $x \neq0$. In that case, the first solution has algebraic and geometric multiplicity $2$, the second algebraic and geometric multiplicity $1$.
ADDED after more thoughts about the second proposition: The second proposition is true. It is easy to see that
$$\det[A]=f(Y)=Y^3 + 2x_{12}x_{13}x_{23} - Y(x^2_{12} + x^2_{13} + x^2_{23})$$.
In order for $f(Y)=0$ to have a root with multiplicity > 1, it has to be also a root of $f'(Y)=0$, which means
$$f'(Y)=3Y^2-(x^2_{12} + x^2_{13} + x^2_{23})=0,$$
which implies
$$Y=\pm\sqrt{\frac{x^2_{12} + x^2_{13} + x^2_{23}}3}.$$
Putting this into $f(Y)=0$ we get
$$0 = f(Y)=Y(Y^2-(x^2_{12} + x^2_{13} + x^2_{23})) + 2x_{12}x_{13}x_{23}=-\frac23(x^2_{12} + x^2_{13} + x^2_{23})Y + 2x_{12}x_{13}x_{23}$$
which implies
$$x_{12}x_{13}x_{23} = \frac13(x^2_{12} + x^2_{13} + x^2_{23})Y.$$
Squaring both sides yields to
$$(x_{12}x_{13}x_{23})^2=\frac19(x^2_{12} + x^2_{13} + x^2_{23})^2Y^2 = \frac19(x^2_{12} + x^2_{13} + x^2_{23})^2\frac{x^2_{12} + x^2_{13} + x^2_{23}}3 =\frac1{27}(x^2_{12} + x^2_{13} + x^2_{23})^3$$
This is the same condition as Yves Daoust arrived in his answer.
Now if we assume that all the $x_{ij}$ are real numbers, then this condition indeed implies that all $x_{ij}$ must be the same. That's because, as every $x_{ij}$ appears squared, we can generally assume that all $x_{ij} \ge 0$. The one can apply the theorem of the inequality of quadratic and geometric mean:
$$\sqrt[3]{x_{12}x_{13}x_{23}} \le \sqrt\frac{x^2_{12} + x^2_{13} + x^2_{23}}{3}$$
Putting this inequality to the 6th power leads to
$$(x_{12}x_{13}x_{23})^2\le \frac1{27}(x^2_{12} + x^2_{13} + x^2_{23})^3$$
and we know that equality (as required for a solution for $Y$ with multiplicity > 1) only happens if $x_{12}=x_{13}=x_{23}$.