How to evaluate the given determinant 
Question Statement:-
  Show that $$\begin{vmatrix}
(b+c)^2 & c^2 & b^2 \\
c^2 & (c+a)^2 & a^2 \\
b^2 & a^2 & (a+b)^2 \\
\end{vmatrix}=2(ab+bc+ca)^3$$


Attempt at a Solution:-
1st attempt(which was in vain):-
LHS:-$$\begin{align}\begin{vmatrix}
(b+c)^2 & c^2 & b^2 \\
c^2 & (c+a)^2 & a^2 \\
b^2 & a^2 & (a+b)^2 \\
\end{vmatrix}=\begin{vmatrix}
2bc & c^2 & b^2 \\
(c^2-a^2)-(c+a)^2 & (c+a)^2 & a^2 \\
(b^2-a^2)-(b
+a)^2 & a^2 & (a+b)^2 \\
\end{vmatrix}\left[\begin{array}{11}
C_1\rightarrow C_1-C_2-C_3\end{array}\right]
=\begin{vmatrix}
2bc & c^2 & b^2 \\
-2(a^2+ac) & (c+a)^2 & a^2 \\
-2(a^2+ab) & a^2 & (a+b)^2 \\
\end{vmatrix}=
2\begin{vmatrix}
bc & c^2 & b^2 \\
-(a^2+ac) & (c+a)^2 & a^2 \\
-(a^2+ab) & a^2 & (a+b)^2 \\
\end{vmatrix}=
\dfrac{2}{bc}\begin{vmatrix}
b^2c^2 & c^2 & b^2 \\
-bc(a^2+ac) & (c+a)^2 & a^2 \\
-bc(a^2+ab) & a^2 & (a+b)^2 \\
\end{vmatrix}\left[C_1\rightarrow bc\cdot C_1\right]=
\dfrac{2}{bc}\begin{vmatrix}
b^2c^2-b^2c^2 & c^2 & b^2 \\
-bc(a^2+ac)-b^2(c+a)^2 & (c+a)^2 & a^2 \\
-bc(a^2+ab)-a^2b^2 & a^2 & (a+b)^2 \\
\end{vmatrix}[C_1\rightarrow C_1-b^2C_2]=
\dfrac{2}{bc}\begin{vmatrix}0 & c^2 & b^2 \\
-b(a+c)(ab+bc+ac) & (c+a)^2 & a^2 \\
-ab(ab+bc+ac) & a^2 & (a+b)^2 \\
\end{vmatrix}\\
=-2\left(\dfrac{ab+bc+ac}{c}\right)\begin{vmatrix}
0 & c^2 & b^2 \\
a+c & (c+a)^2 & a^2 \\
a & a^2 & (a+b)^2 \\
\end{vmatrix}
\end{align}$$
I was pretty much stuck after this so I tried another approach.
2nd Attempt:-
$$\begin{vmatrix}
(b+c)^2 & c^2 & b^2 \\
c^2 & (c+a)^2 & a^2 \\
b^2 & a^2 & (a+b)^2 \\
\end{vmatrix}=
(abc)^4\begin{vmatrix}
\left(\dfrac{1}{b}+\dfrac{1}{c}\right)^2 & \dfrac{1}{b^2} & \dfrac{1}{c^2} \\
\dfrac{1}{a^2} & \left(\dfrac{1}{a}+\dfrac{1}{c}\right)^2 & \dfrac{1}{c^2} \\
\dfrac{1}{a^2} & \dfrac{1}{b^2} & \left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2 \\
\end{vmatrix}\left[\begin{array}{11}
R_1\rightarrow \dfrac{R_1}{b^2c^2} \\
R_2\rightarrow \dfrac{R_2}{a^2c^2} \\
R_2\rightarrow \dfrac{R_3}{a^2b^2 }\end{array}\right]$$
And after starting along the route that I have shown in the second attempt I figured it was much more useless than the previous one. So, I thought that the Mathematics Stack Exchange is the only route left. So, your help is very much needed. 

Let us continue with your valiant first attempt (note that you incorrectly missed out a factor of $2$, from your third stage to the fourth stage)
$$-\color{red}{2}\left(\dfrac{ab+bc+ac}{c}\right)\begin{vmatrix}
0 & c^2 & b^2 \\
a+c & (c+a)^2 & a^2 \\
a & a^2 & (a+b)^2 \\\end{vmatrix}$$
Using a brute force approach to evaluate the determinant of the $3\times 3$ matrix (aided by the fact that the top left entry is $0$), the determinant evaluates to
$$0-c^2((a+b)^2(a+c)-a^2)+b^2(a^2(a+c)-(c+a)^2a)$$
which, after a fair amount of algebraic manipulation, simplifies to
$$-c(a^2c^2+b^2c^2+a^2b^2+2a^2bc+2abc^2+2ab^2c)=-c(ab+bc+ac)^2$$
We thus have the desired result
$$-2\left(\frac{ab+bc+ac}{c}\right)\times-c(ab+bc+ac)^2=2(ab+bc+ac)^3$$ 

A less brutal approach is to make use of the Matrix determinant lemma see here. 
The matrix, denote by $M$, can be expressed as
$$M=vv^T-A=-(A+(-vv^T))$$
where $v= \left( \begin{array}{c}
b+c  \\
a+c \\
a+b \end{array} \right) $ and $A= (ab+bc+ca)\left( \begin{array}{ccc}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0 \end{array} \right) $
Invoking the  Matrix determinant lemma, we have
$$\det(M)=-(1-v^TA^{-1}v)\det(A)\tag{Eq.1}$$
Now, it is straightforward to show that 
$$\det(A)=2(ab+bc+ac)^3$$ and $$A^{-1}=\frac{1}{2(ab+bc+ca)}\left( \begin{array}{ccc}
-1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 1 & -1 \end{array} \right)$$
Leading to $v^TA^{-1}v=2$
Substituting all the values into Eq.$1$, we have
$$\det(M)=-(1-2)\times2(ab+bc+ac)^3=2(ab+bc+ca)^3$$
