# Prove that:

Question:

Prove that:

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

Solution:

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

$=\left|\begin{array}{ccc}a & b & c \\ a-b & b-c & c-a \\ a+b+c & c+a+b & a+b+c\end{array}\right| \quad$ [Applying $R_{3} \rightarrow R_{3}+R_{2}$ ]

$=\left(\begin{array}{c}a+b+c \\ \end{array}\right)\left|\begin{array}{ccc}a & b & c \\ a-b & b-c & c-a \\ 1 & 1 & 1\end{array}\right| \quad$ [Taking $(\mathrm{a}+\mathrm{b}+\mathrm{c})$ common]

$=(a+b+c)\left|\begin{array}{lll}a & b & c \\ b & c & a \\ 1 & 1 & 1\end{array}\right|$    [Applying $R_{2} \rightarrow R_{1}-R_{2}$ ]

$=(a+b+c)\left|\begin{array}{ccc}a-b & b-c & c \\ b-c & c-a & a \\ 0 & 0 & 1\end{array}\right| \quad\left[C_{1} \rightarrow C_{1}-C_{2}\right.$ and $\left.C_{2} \rightarrow C_{2}-C_{3}\right]$

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

$=(a+b+c)\left[-\left\{a c-b c-a^{2}+a b-b^{2}-c^{2}+2 b c\right\}\right]$

$=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$

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