Using the property of determinants and without expanding,prove that:
$\left|\begin{array}{lll}b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y\end{array}\right|=2\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|$

(N/A) Let $\Delta = \left|\begin{array}{lll}b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y\end{array}\right|$.
Applying $R_1 \rightarrow R_1 + R_2 + R_3$,we get:
$\Delta = \left|\begin{array}{lll}2(a+b+c) & 2(p+q+r) & 2(x+y+z) \\ c+a & r+p & z+x \\ a+b & p+q & x+y\end{array}\right| = 2 \left|\begin{array}{lll}a+b+c & p+q+r & x+y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y\end{array}\right|$.
Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$,we get:
$\Delta = 2 \left|\begin{array}{lll}a+b+c & p+q+r & x+y+z \\ -b & -q & -y \\ -c & -r & -z\end{array}\right|$.
Applying $R_1 \rightarrow R_1 + R_2 + R_3$,we get:
$\Delta = 2 \left|\begin{array}{lll}a & p & x \\ -b & -q & -y \\ -c & -r & -z\end{array}\right|$.
Taking $-1$ common from $R_2$ and $R_3$,we get:
$\Delta = 2(-1)(-1) \left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right| = 2 \left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|$.
Hence,the result is proved.