Question

If $\left|\begin{array}{ccc}a+b+2c& a& b\\ c& 2a+b+c& b\\ c& a& a+2b+c\end{array}\right|=2$, 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)

$\left|\begin{array}{ccc}a+b+2c& a& b\\ c& 2a+b+c& b\\ c& a& a+2b+c\end{array}\right|=2$
$\Rightarrow 2(a+b+c)\left|\begin{array}{ccc}1& a& b\\ 1& b+c+2a& b\\ 1& a& c+a+2b\end{array}\right|=2$
[ Applying $C_1\to C_1+C_2+C_3$ and
taking $2(a+b+c)$ common from $C_1]$
$\Rightarrow 2(a+b+c)\left|\begin{array}{ccc}1& a& b\\ 0& b+c+a& 0\\ 0& 0& c+a+b\end{array}\right|=2$
[ Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1]$
$\Rightarrow 2(a+b+c{)}^3=2$ [ expanding along $C_1]$
$\Rightarrow (a+b+c{)}^3=1$
$\Rightarrow a+b+c=1$
Now, $a^3+b^3+c^3-3abc$
$=(a+b+c)\left(a^2+b^2+c^2-ab-bc-ca\right)$
$=1\cdot \left[a^2+b^2+c^2-ab-bc-ca\right]$
$=(a+b+c{)}^2-2ab-2bc-2ca-ab-bc-ca$
$=1-2ab-2bc-2ca-ab-bc-ca$
$=1-3ab-3bc-3ca$