Question

The value of $\left|\begin{array}{ccc}a& b+c& a^3\\ b& c+a& b^3\\ c& a+b& c^3\end{array}\right|$ is

• A. $(a-b)(b-c)(c-a)$
• B. $abc(a-b)(b-c)(c-a)$
• C. $(a+b+c{)}^2(a-b\left)\right(b-c\left)\right(a-c)$
• D. none of these

Answer 1

The correct option is C $(a+b+c{)}^2(a-b\left)\right(b-c\left)\right(a-c)$

$\left|\begin{array}{ccc}a& b+c& a^3\\ b& c+a& b^3\\ c& a+b& c^3\end{array}\right|$
$C_2\to C_2+C_1$
$=\left|\begin{array}{ccc}a& a+b+c& a^3\\ b& a+b+c& b^3\\ c& a+b+c& c^3\end{array}\right|$
$=(a+b+c)\left|\begin{array}{ccc}a& 1& a^3\\ b& 1& b^3\\ c& 1& c^3\end{array}\right|$
$R_1\to R_1-R_2,R_2\to R_2-R_3$
$=(a+b+c)\left|\begin{array}{ccc}a-b& 0& a^3-b^3\\ b-c& 0& b^3-c^3\\ c& 1& c^3\end{array}\right|$
$=(a+b+c)(a-b)(b-c)\left|\begin{array}{ccc}1& 0& a^2+b^2+ab\\ 1& 0& b^2+c^2+bc\\ c& 1& c^3\end{array}\right|$
$=(a+b+c)(a-b)(b-c)[c^2-a^2+b(c-a\left)\right]$
$=(a+b+c{)}^2(a-b\left)\right(b-c\left)\right(c-a)$