# Using properties of determinants, prove the following : $$\begin{bmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{bmatrix} = 2(a+b+c)^3.$$

Toolbox:
• If each element of a row (or column) of a determinant is multiplied by k,then its value is also multiplied by k.
• By this property we can take out any common factor from any one row or any column of the given determinant.
Step 1:
Let $\Delta=\begin{vmatrix}a+b+2c & a & b\\c &b+c+2a & b\\c & a & c+a+2b\end{vmatrix}$
Let us take $C_1,C_2$ and $C_3$,hence we apply $C_1\rightarrow C_1+C_2+C_3$
$\Delta=\begin{vmatrix}2a+2b+2c & a & b\\2a+2b+2c &b+c+2a & b\\2a+2b+2c & a & c+a+2b\end{vmatrix}$
Take $2(a+b+c)$ as a common factor from $C_1$
$\Delta=2(a+b+c)\begin{vmatrix}1 & a & b\\1 &b+c+2a & b\\1 & a & c+a+2b\end{vmatrix}$
Step 2:
Now by applying $R_1\rightarrow R_1-R_2$ and $R_2\rightarrow R_2-R_3$
$\Delta=2(a+b+c)\begin{vmatrix}0 & -(a+b+c) & 0\\0 &a+b+c & -(a+b+c)\\1 & a & c+a+2b\end{vmatrix}$
By applying $R_1\rightarrow R_1+R_2$
$\Delta=2(a+b+c)\begin{vmatrix}0 & 0 & -(a+b+c)\\0 &a+b+c & -(a+b+c)\\1 & a & c+a+2b\end{vmatrix}$
Step 3:
Now expanding along $R_1$ we get
$\Delta=2(a+b+c)[-(a+b+c)\big(0\times a-1\times (a+b+c)\big)]$
$\;\;\;=2(a+b+c)(a+b+c)^2$
$\;\;\;=2(a+b+c)^3$
Hence $\begin{vmatrix}a+b+2c & a & b\\c &b+c+2a & b\\c & a & c+a+2b\end{vmatrix}=2(a+b+c)^3$
Hence proved.