<s>[INST] Computing determinant without expansion $$\begin{align}\mathrm D &= \left|\begin{matrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (a+c)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{matrix}\right|\\ &= (a+b+c)\left|\begin{matrix} b+c - a & a^2 & a^2 \\ b - a -c & (a+c)^2 & b^2 \\ 0 & c^2 & (a+b)^2 \end{matrix}\right| \\ &= (a+b+c)^2\left|\begin{matrix} b+c - a & 0 & a^2 \\ b - a -c & a+c - b & b^2 \\ 0 & c - a-b & (a+b)^2 \end{matrix}\right|\\ &= (a+b+c)^2\left|\begin{matrix} b+c - a & 0 & a^2 \\ 0 & a+c - b & b^2 \\ c - a-b & c - a-b & (a+b)^2 \end{matrix}\right|\end{align}$$
Can $\rm D$ be further simplified without expanding ?  I feel it should be because this was competition question.
 [/INST] We already reduced the problem to calculate 
$$D’=\left|\begin{matrix}
b+c - a & 0 & a^2 \\
0 & a+c - b & b^2 \\
b & a & -ab 
\end{matrix}\right|$$
If $a=0$ then 
$$D’=\left|\begin{matrix}
b+c & 0 & 0\\
0 & c - b & b^2 \\
b & 0 & 0 
\end{matrix}\right|=0.$$
If $b=0$ then 
$$D’=\left|\begin{matrix}
c - a & 0 & a^2 \\
0 & a+c & 0 \\
0 & a & 0 
\end{matrix}\right|=0.$$
Otherwise put $R’_1=R_1+\frac abR_3$ and $R’_2=R_2+\frac baR_3$. Then 
$$D’=\left|\begin{matrix}
b+c & \frac {a^2}b & 0\\
\frac {b^2}a  & a+c & 0 \\
b & a & -ab 
\end{matrix}\right|=-ab\left|\begin{matrix}
b+c & \frac {a^2}b \\
\frac {b^2}a  & a+c \\
\end{matrix}\right|=-ab[(a+c)(b+c)-ab]=-ab[ac+bc+c^2]=-abc(a+b+c).$$
The latter formula holds also when $a=0$ or $b=0$. Finally, 
$$D=(a+b+c)(-2)D’=2(a+b+c)^3abc.$$
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