Question

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

• A. 4 abc
• B. 6 abc
• C. 8 abc
• D. 2 abc

Answer 1

The correct option is C 8 abc

Given,

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

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

$=(b+c)(2ba+2ca)-(a-c)(2b^2-2bc)+(a-b)(-2c^2+2bc)$

$=8abc$