Determinant of a matrix without actually expanding it. Given a Matrix $$A =  \begin{vmatrix}
a^2 & (s-a)^2 & (s-a)^2\\
(s-b)^2 & b^2 & (s-b)^2 \\
(s-c)^2 & (s-c)^2 & c^2\\ 
\end{vmatrix}$$
and $$a+b+c = 2s$$ , how can I find its determinant without expanding?
I've tried finding all factors which make two columns (/rows) similar or proportional. Only with s=0 all the rows and columns becomes 1 (after taking out the common value). With s=a $$\left|\matrix{
a^2 & 0 & 0\\
(a-b)^2=c^2 & b^2 & (a-b)^2=c^2 \\
(a-c)^2=b^2 & (a-c)^2=b^2 & c^2\\ 
}\right|=0$$
Here is a list of trivial factors
(s-a), (s-b), (s-c), (s-0). I'm given a hint that the determinant is a degree 6 polynomial, which I don't understand cause max degree of s is 4, please do explain how degree is 6.
I think det(A) = (s-a)(s-b)(s-c)(s-0).
Are there any more factors? What are they?
Edit: Please also explain how to arrive at the solution.
 A: The equation $a+b+c=2s$ has a geometric interpretation: $s$ is the semiperimeter of a triangle of sides $a,\,b,\,c$. If this triangle is degenerate, say with $a=b+c$, then $s=a$ and$$A=\left|\begin{array}{ccc}
\left(b+c\right)^{2} & 0 & 0\\
c^{2} & b^{2} & c^{2}\\
b^{2} & b^{2} & c^{2}
\end{array}\right|=0,$$so $(s-a)(s-b)(s-c)|A$. A similar treatment of $-a=b+c\implies s=0$ can prove $s|A$. Since $A$ is of dimension $6$ and is homogeneous when all sides are rescaled, the symmetries require a result proportional to $s^3(s-a)(s-b)(s-c)$. An equilateral triangle with $a=b=c=2,\,s=3,\,A=54$ proves the coefficient is $2$, as in @JanEerland's answer.
A: 
Not a 'real' answer, but it was too big for a comment.

I wrote and ran some Mathematica-code:
In[1]:=Clear["Global`*"];
FullSimplify[
 Det[{{a^2, (s - a)^2, (s - a)^2}, {(s - b)^2, 
    b^2, (s - b)^2}, {(s - c)^2, (s - c)^2, c^2}}], 
 Assumptions -> a + b + c == 2 s]

Running the code gives:
Out[1]=-2 (b - s) (b + c - s) s^3 (-c + s)

If you're not only looking for the solution, it is not very hard to show that this is correct.
