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

Answer: (A)

$1 - 3ab - 3bc - 3ca$

Answer 2

$\begin{vmatrix}a+b+2 c & a & b \\\ c & 2 a+b+c & b \\\ c & a & a+2 b+c\end{vmatrix} =2$
$\Rightarrow 2(a+b+c)\begin{vmatrix} 1 & a & b \\\ 1 & b+c+2 a & b \\\ 1 & a & c+a+2 b\end{vmatrix}=2$
[ 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, $a^{3}+b^{3}+c^{3}-3 a b c$
$=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
$= 1 \cdot\left[a^{2}+b^{2}+c^{2}-a b-b c-c a\right]$
$= (a+b+c)^{2}-2 a b-2 b c-2 c a-a b-b c-c a$
$= 1-2 a b-2 b c-2 c a-a b-b c-c a$
$= 1-3 a b-3 b c-3 c a$