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Using properties of determinants prove that:
$\left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right|=3(a+b+c)(a b+b c+c a)$
$\left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right|=3(a+b+c)(a b+b c+c a)$
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Verified Answer
Let $\Delta=\left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right|$
Applying $\mathrm{C}_1 \rightarrow \mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3$
$\Delta=\left|\begin{array}{ccc}
\mathrm{a}+\mathrm{b}+\mathrm{c} & -\mathrm{a}+\mathrm{b} & -\mathrm{a}+\mathrm{c} \\
\mathrm{a}+\mathrm{b}+\mathrm{c} & 3 \mathrm{~b} & -\mathrm{b}+\mathrm{c} \\
\mathrm{a}+\mathrm{b}+\mathrm{c} & -\mathrm{c}+\mathrm{b} & 3 \mathrm{c}
\end{array}\right|$
Taking common $(\mathrm{a}+\mathrm{b}+\mathrm{c})$ from $\mathrm{C}_1$
$\Delta=(\mathrm{a}+\mathrm{b}+\mathrm{c})\left|\begin{array}{ccc}
1 & -\mathrm{a}+\mathrm{b} & -\mathrm{a}+\mathrm{c} \\
1 & 3 \mathrm{~b} & -\mathrm{b}+\mathrm{c} \\
1 & -\mathrm{c}+\mathrm{b} & 3 \mathrm{c}
\end{array}\right|$
Applying $\mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1$ and $\mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1$
$\Delta=(\mathrm{a}+\mathrm{b}+\mathrm{c})\left|\begin{array}{ccc}
1 & -\mathrm{a}+\mathrm{b} & -\mathrm{a}+\mathrm{c} \\
0 & 2 \mathrm{~b}+\mathrm{a} & \mathrm{a}-\mathrm{b} \\
0 & \mathrm{a}-\mathrm{c} & 2 \mathrm{c}+\mathrm{a}
\end{array}\right|$
Expanding along $\mathrm{C}_1$, we get
$\Delta=3(\mathrm{a}+\mathrm{b}+\mathrm{c})(\mathrm{ab}+\mathrm{ba}+\mathrm{ca})$
Applying $\mathrm{C}_1 \rightarrow \mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3$
$\Delta=\left|\begin{array}{ccc}
\mathrm{a}+\mathrm{b}+\mathrm{c} & -\mathrm{a}+\mathrm{b} & -\mathrm{a}+\mathrm{c} \\
\mathrm{a}+\mathrm{b}+\mathrm{c} & 3 \mathrm{~b} & -\mathrm{b}+\mathrm{c} \\
\mathrm{a}+\mathrm{b}+\mathrm{c} & -\mathrm{c}+\mathrm{b} & 3 \mathrm{c}
\end{array}\right|$
Taking common $(\mathrm{a}+\mathrm{b}+\mathrm{c})$ from $\mathrm{C}_1$
$\Delta=(\mathrm{a}+\mathrm{b}+\mathrm{c})\left|\begin{array}{ccc}
1 & -\mathrm{a}+\mathrm{b} & -\mathrm{a}+\mathrm{c} \\
1 & 3 \mathrm{~b} & -\mathrm{b}+\mathrm{c} \\
1 & -\mathrm{c}+\mathrm{b} & 3 \mathrm{c}
\end{array}\right|$
Applying $\mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1$ and $\mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1$
$\Delta=(\mathrm{a}+\mathrm{b}+\mathrm{c})\left|\begin{array}{ccc}
1 & -\mathrm{a}+\mathrm{b} & -\mathrm{a}+\mathrm{c} \\
0 & 2 \mathrm{~b}+\mathrm{a} & \mathrm{a}-\mathrm{b} \\
0 & \mathrm{a}-\mathrm{c} & 2 \mathrm{c}+\mathrm{a}
\end{array}\right|$
Expanding along $\mathrm{C}_1$, we get
$\Delta=3(\mathrm{a}+\mathrm{b}+\mathrm{c})(\mathrm{ab}+\mathrm{ba}+\mathrm{ca})$
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