Question

Prove that :

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

Answer 1

$$\begin{gathered} ∆=\left|\begin{array}{ccc}a& b& c\\ a-b& b-c& c-a\\ b+c& c+a& a+b\end{array}\right| \\ =\left|\begin{array}{ccc}a& b& c\\ a-b& b-c& c-a\\ a+b+c& c+a+b& a+b+c\end{array}\right|\left[\mathrm{Applying}{R}_3\to R_3+R_2\right] \\ =(a+b+c)\left|\begin{array}{ccc}a& b& c\\ a-b& b-c& c-a\\ 1& 1& 1\end{array}\right|\left[\mathrm{Taking}(\mathrm{a}+\mathrm{b}+\mathrm{c})\mathrm{common}\right] \\ =(a+b+c)\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ 1& 1& 1\end{array}\right|\left[\mathrm{Applying}{R}_2\to R_1-R_2\right] \\ =(a+b+c)\left|\begin{array}{ccc}a-b& b-c& c\\ b-c& c-a& a\\ 0& 0& 1\end{array}\right|\left[C_1\to C_1-C_2\mathrm{and}{C}_2\to C_2-C_3\right] \\ =(a+b+c)\left[-1\left\{\left(a-b\right)\left(c-a\right)-{\left(b-c\right)}^2\right\}\right] \\ =(a+b+c)\left[-\left\{ac-bc-a^2+ab-b^2-c^2+2bc\right\}\right] \\ =(a+b+c)\left(a^2+b^2+c^2-ab-bc-ca\right) \\ =a^3+b^3+c^3-3abc \end{gathered}$$