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# Using properties of determinants, prove that : $\begin{vmatrix} a+b+nc & (n-1)a & (n-1)b \\ (n-1)c & b+c+na & (n-1)b \\ (n-1)c & (n-1)a & c+a+nb \end{vmatrix} = n(a+b+c)^3$

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A)
Toolbox:
• Elementary transformations can be made in a determinant
• a)by interchanging two rows or columns.
• b)Two or more columns can be added or subtracted.
• The value of the determinant can be obtained by expanding along any rows or columns
Step 1:
Let $\Delta= \begin{vmatrix} a+b+nc & (n-1)a & (n-1)b \\ (n-1)c & b+c+na & (n-1)b \\ (n-1)c & (n-1)a & c+a+nb \end{vmatrix}$
Apply : $C_1 \to C_1+C_2+C_3$
$\Delta= \begin{vmatrix} na+nb+nc & (n-1)a & (n-1)b \\ na+nb+nc & b+c+na & (n-1)b \\ na+nb+nc & (n-1)a & c+a+nb \end{vmatrix}$
Take $n(a+b+c)$ as the common factor from $C_1$
$\Delta= \begin{vmatrix} 1 & (n-1)a & (n-1)b \\ 1 & (b+c+na) & (n-1)b \\ 1 & (n-1)a & c+a+nb \end{vmatrix}$
Apply $R_2 \to R_2-R_1\; and \; R_3\to R_3-R_1$
$\Delta=n(a+b+c)\begin{vmatrix} 1 & (n-1)a & (n-1)b \\ 0 & a+b+c & a+b+c \\ 0 & 0 & a+b+c \end{vmatrix}$
Take $(a+b+c)$ as common factor from $R_2\;and\;R_3$
$\Delta=n(a+b+c)^3 \begin{vmatrix} 1 & (n-1)a & (n-1)b \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{vmatrix}$
Step 2:
Expanding along $R_1$ we get
$\Delta=n(a+b+c)^3 \bigg[1-(1-0)-0\bigg]$
$\Delta=n(a+b+c)^3$
Hence proved