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

To prove:- $\left|\begin{array}{ccc}b+c& a-b& a\\ c+a& b-c& b\\ a+b& c-a& c\end{array}\right|=3abc-a^3-b^3-c^3$

Proof:-

Taking L.H.S.-

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

Applying $R_1\to R_1+R_2+R_3$

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

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

Now, expanding along $R_1$, we have

$=(a+b+c)[2(bc-c^2-bc+ab)-0+1(c^2-a^2-ab-b^2+ac+bc)]$

$=(a+b+c)(2ab-2c^2+c^2-a^2-b^2-ab+ac+bc)$

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

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

$=-(a^3+b^3+c^3-3abc)$

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

$=$ R.H.S.

Hence proved