Question

Using properties of determinant, prove that $\left|\begin{array}{ccc}b+c& a-b& a\\ c+a& b-c& b\\ a+b& c-a& c\end{array}\right|=3abc-a^3-b^3-c^3$

Answer 1

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

$R_1\to R_1+R_2+R_3$

$=\left|\begin{array}{ccc}2(a+b+c)& 0& a+b+c\\ c+a& b-c& b\\ a+b& c-a& c\end{array}\right|$

Taking $a+b+c$ common from $R_1$

$=(a+b+c)\left|\begin{array}{ccc}2& 0& 1\\ c+a& b-c& b\\ a+b& c-a& c\end{array}\right|$

$C_1\to C_1-2C_3$

$=(a+b+c)\left|\begin{array}{ccc}0& 0& 1\\ c+a-2b& b-c& b\\ a+b-2c& c-a& c\end{array}\right|$

$R_1\to R_1+R_2$

$=(a+b+c)\left|\begin{array}{ccc}0& 0& 1\\ c+a-2b+b-c& b-c& b\\ a+b-2c+c-a& c-a& c\end{array}\right|$

$=(a+b+c)\left|\begin{array}{ccc}0& 0& 1\\ a-b& b-c& b\\ b-c& c-a& c\end{array}\right|$

$=(a+b+c)[0-0+1(a-b)(c-a)-{(b-c)}^2]$

$=(a+b+c)[ac-a^2-bc+ba-b^2-c^2+2bc]$

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

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

$=-(a^3+b^3+c^3-3abc)$ by using the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ba-bc-ca)$

$=3abc-a^3-b^3-c^3$

Hence proved.