Question

$\left|\begin{array}{ccc}b+c& a-b& a\\ c+a& b-c& b\\ a+b& c-a& c\end{array}\right|=$

• A. $a^3+b^3+c^3-3abc$
• B. $3abc-a^3-b^3-c^3$
• C. $a^3+b^3+c^3-a^{2}b-b^{2}c-c^{2}a$
• D. $(a+b+c)(a^{2}b-b^{2}c-c^2+ab+bc+ca)$

Answer 1

The correct option is B $3abc-a^3-b^3-c^3$

$\Delta =\left|\begin{array}{ccc}2(a+b+c)& 0& a+b+c\\ c+a& b-c& b\\ a+b& c-a& c\end{array}\right|by{R}_1\to R_1+R_2+R_3\Delta =(a+b+c)\left|\begin{array}{ccc}2& 0& 1\\ c+a& b-c& b\\ a+b& c-a& c\end{array}\right|$
On expanding, $-(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$=-(a^3+b^3+c^3-3abc)=3abc-a^3-b^3-c^3$.
Trick : Put a=1, b=2, c=3 and check it.