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. 
 A: 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
$$\det(M)=-(1-2)\times2(ab+bc+ac)^3=2(ab+bc+ca)^3$$
