Proving the determinant of a $3\times 3$ matrix is given by $2s^3(s-a)(s-b)(s-c)$. Prove:

$$\begin{vmatrix}a^2&(s-a)^2&(s-a)^2\\(s-b)^2&b^2&(s-b)^2\\(s-c)^2&(s-c)^2&c^2\end{vmatrix}=2s^3(s-a)(s-b)(s-c),\;\;s=\frac{a+b+c}{2}$$

My attempt:
Let $c_1,c_2,c_3$ be columns and $r_1,r_2,r_3$ be rows.
$c_3-c_2:$
$$\begin{vmatrix}a^2&(s-a)^2&0\\(s-b)^2&b^2&s(s-2b)\\(s-c)^2&(s-c)^2&-s(s-2c)\end{vmatrix}$$
$c_2-c_1:$
$$\begin{vmatrix}a^2&s(s-2a)&0\\(s-b)^2&-s(s-2b)&s(s-2b)\\(s-c)^2&0&-s(s-2c)\end{vmatrix}$$
$c_2\cdot\frac{1}{s},c_3\cdot\frac{1}{s}$
$$s^2\begin{vmatrix}a^2&(s-2a)&0\\(s-b)^2&-(s-2b)&(s-2b)\\(s-c)^2&0&-(s-2c)\end{vmatrix}$$
Then, by LaPLace on the $3^{\text{rd}}$ column, I got:
$$-(s-2b)(s-2a)(s-c)^2-(s-2c)\left(a^2(2b-s)-(s-2a)(s-b)^2\right)$$
Which got too complicated. May I ask how to end this task?
 A: COMMENT.-Put $S_a=s-a$ and analogues you have
$$\begin{vmatrix}a^2&S_a^2&S_a^2\\S_b^2&b^2&S_b^2\\S_c^2&S_c^2&c^2\end{vmatrix}=2s^3S_aS_bS_c$$ By property of determinants one has
$$LHS=\begin{vmatrix}a^2&S_a^2&0\\S_b^2&b^2&S_b^2-b^2\\S_c^2&S_c^2&S_c-c^2\end{vmatrix}=2S_a^2S_b^2S_c^2-a^2S_b^2S_c^2-b^2S_a^2S_c^2-c^2S_a^2S_b^2$$ With this the verification of the identity is straightforward but tedious.
An easier way is to consider $a,b,c$ as the sides of a triangle so you have to prove
$$LHS=R^6\begin{vmatrix}4(\sin \alpha)^2&(-\sin \alpha+\sin\beta+\sin \gamma)^2&(-\sin \alpha+\sin\beta+\sin \gamma)^2\\(\sin \alpha-\sin\beta+\sin \gamma)^2&4(\sin \beta)^2&(\sin \alpha-\sin\beta+\sin \gamma)^2\\(\sin \alpha+\sin\beta-\sin \gamma)^2&(\sin \alpha+\sin\beta-\sin \gamma)^2&4(\sin \gamma)^2\end{vmatrix}$$ where $R$ is the radius of the circumscribed circle.
You have for the $RHS$ $$s=R(\sin \alpha+\sin \beta+\sin \gamma)\\s-a=R(-\sin \alpha+\sin\beta+\sin \gamma)\\s-b=R(\sin \alpha-\sin\beta+\sin \gamma)\\s-c=R(\sin \alpha+\sin\beta-\sin \gamma)$$ This way is easier without doubt.
