Question

If $\begin{vmatrix}a+b+2c&a&b\\ c&2a+b+c&b\\ c&a&a+2b+c\end{vmatrix} = 2 $, then $a^3 + b^3 + c^3 - 3abc$ =

• A. $1 - 3ab - 3bc - 3ca$
• B. $0$
• C. $1 - 2ab - 2bc - 2ca$
• D. $1$

Answer 1

The Correct Option is A

$\begin{vmatrix}a+b+2 c & a & b \\ c & 2 a+b+c & b \\ c & a & a+2 b+c\end{vmatrix} =2$
$\Rightarrow 2(a+b+c)\begin{vmatrix} 1 & a & b \\ 1 & b+c+2 a & b \\ 1 & a & c+a+2 b\end{vmatrix}=2 $
[ Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3} $ and
taking $2( a + b + c) $ common from $C_{1} ] $
$\Rightarrow 2(a+b+c) \begin{vmatrix}1 & a & b \\ 0 & b+c+a & 0 \\ 0 & 0 & c+a+b\end{vmatrix} =2$
[ Applying $ R_{2} \rightarrow R_{2}-R_{1}$ and $ R_{3} \rightarrow R_{3}-R_{1}] $
$\Rightarrow 2(a+b+c)^{3}=2$ [ expanding along $ C_{1}] $
$\Rightarrow (a+b+c)^{3}=1 $
$\Rightarrow a+b+c=1 $
Now, $ a^{3}+b^{3}+c^{3}-3 a b c $
$ =(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) $
$ = 1 \cdot\left[a^{2}+b^{2}+c^{2}-a b-b c-c a\right] $
$= (a+b+c)^{2}-2 a b-2 b c-2 c a-a b-b c-c a $
$ = 1-2 a b-2 b c-2 c a-a b-b c-c a $
$ = 1-3 a b-3 b c-3 c a$