Question

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

Answer 1

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

Taking L.H.S

$=(b+c)\left[\right(bc+c^2-bc-ab\left)\right]-(a+b)(c^2+ac-ab-b^2)+a(c^2+a^2+2a-ab-b^2-ac-bc)$

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

$=b{c}^2-a{b}^2+c^3-abc-[a{c}^2+a^{2}c-a^{2}b-a{b}^2+b{c}^2+abc-a{b}^2+b^3]+a{c}^2+a^3-a{b}^2+a^{2}c-a^{2}b-abc$

$=b{c}^2-a{b}^2+c^3-abc-a{c}^2+a^{2}c+a^{2}b+a{b}^2-b{c}^2-abc+a{b}^2+b^3+a{c}^2+a^3-a{b}^2+a^{2}c+a^{2}b-abc$

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