#### Explain solution RD Sharma class 12 chapter Determinants exercise multiple choise question 24 maths

Correct option (c)

Hint:

Using properties of determinant.

Given:

$Let\: \:\;\Delta =\begin{vmatrix} a-b &b+c &a \\ b-c &c+a &b \\ c-a &a+b &c \end{vmatrix}$

We have to find the value of $\Delta$.

Solution:

$\Delta =\begin{vmatrix} a-b &b+c &a \\ b-c &c+a &b \\ c-a &a+b &c \end{vmatrix}$

Using properties,

$\Rightarrow \Delta =\begin{vmatrix} a &b &a \\ b &c &b \\ c &a &c \end{vmatrix}+\begin{vmatrix} -b &c &a \\ -c &a &b \\ -a &b &c \end{vmatrix}$

Here two columns are identical

$\Rightarrow \begin{vmatrix} a &b &a \\ b &c &b \\ c &a &c \end{vmatrix}=0$

$\Rightarrow \Delta =0+(-1)\begin{vmatrix} b &c &a \\ c &a &b \\ a &b &c \end{vmatrix}$

Applying R1→R1+R2+R3

$\Rightarrow \Delta =-\begin{vmatrix} a+b+c &a+b+c &a+b+c \\ c &a &b \\ a &b &c \end{vmatrix}$

Taking (a+b+c) common from R1

$\Rightarrow \Delta =-(a+b+c)\begin{vmatrix} 1 &1 &1 \\ c &a &b \\ a &b &c \end{vmatrix}$

Expanding along R1, we have

$\Rightarrow \Delta =-(a+b+c)[(ac-b^{2})-(c^{2}-ab)+(bc-a^{2})]$

$\Rightarrow \Delta =-(a+b+c)[ac-b^{2}-c^{2}+ab+bc-a^{2}]$

$\Rightarrow \Delta =(a+b+c)[a^{2}+b^{2}+c^{2}-ab-bc-ca]$

$[\because a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)]$

$\Rightarrow \Delta =a^{3}+b^{3}+c^{3}-3abc$

Hence, this is the required answer.