Question

The value of the determinant $\left|\begin{array}{ccc}a-b& b+c& a\\ b-c& c+a& b\\ c-a& a+b& c\end{array}\right|$ is

(a) $a^3+b^3+c^3$
(b) 3bc
(c) $a^3+b^3+c^3-3abc$
(d) none of these

Answer 1

$$\begin{gathered} \left|\begin{array}{ccc}a-b& b+c& a\\ b-c& c+a& b\\ c-a& a+b& c\end{array}\right| \\ =\left|\begin{array}{ccc}-b& b+c+a& a\\ -c& c+a+b& b\\ -a& a+b+c& c\end{array}\right|\left[\mathrm{Applying}{C}_1\to C_1-C_3\mathrm{and}{C}_2\to C_2+C_3\right] \\ =\left(-1\right)\left(a+b+c\right)\left|\begin{array}{ccc}b& 1& a\\ c& 1& b\\ a& 1& c\end{array}\right|\left[\mathrm{Taking}\left(-1\right)\mathrm{common}\mathrm{from}{C}_1\mathrm{and}\left(a+b+c\right)\mathrm{common}\mathrm{from}{C}_2\right] \\ =\left(-1\right)\left(a+b+c\right)\left|\begin{array}{ccc}b& 1& a\\ c-b& 0& b-a\\ a-b& 0& c-a\end{array}\right|\left[\mathrm{Applying}{R}_2\to R_2-R_1\mathrm{and}{R}_3\to R_3-R_1\right] \\ =\left(-1\right)\left(a+b+c\right)\left[-\left(c-b\right)\left(c-a\right)+\left(b-a\right)\left(a-b\right)\right] \\ =\left(-1\right)\left(a+b+c\right)\left[-c^2+ac+bc-ab+ba-b^2-a^2+ab\right] \\ =\left(-1\right)\left(a+b+c\right)\left(-a^2-b^2-c^2+ab+bc+ac\right) \\ =\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right) \\ =a^3+a{b}^2+a{c}^2-a^{2}b-abc-a^{2}c+b{a}^2+b^3+b{c}^2-a{b}^2-b^{2}c-abc+c{a}^2+c{b}^2+c^3-acb-b{c}^2-a{c}^2 \\ =a^3+b^3+c^3-3abc \end{gathered}$$

Hence, the correct option is (c).