Question

Prove that
\[\begin{vmatrix}- bc & b^2 + bc & c^2 + bc \\ a^2 + ac & - ac & c^2 + ac \\ a^2 + ab & b^2 + ab & - ab\end{vmatrix} = \left( ab + bc + ca \right)^3\]

Answer 1

\[∆ = \begin{vmatrix}- bc & b^2 + bc & c^2 + bc \\ a^2 + ac & - ac & c^2 + ac \\ a^2 + ab & b^2 + ab & - ab\end{vmatrix}\]

\[ = \frac{1}{abc}\begin{vmatrix}- abc & a b^2 + abc & a c^2 + abc \\ a^2 b + abc & - abc & c^2 b + abc \\ a^2 c + abc & b^2 c + abc & - abc\end{vmatrix} \left[\text{ Applying }R_1 \text{ to }aR_1 , R_2 \text{ to }bR_2\text{ and }R_3 \text{ to }cR_3\text{ and then dividing by abc }\right]\]

\[ = \frac{abc}{abc}\begin{vmatrix}- bc & ab + ac & ac + ab \\ ab + bc & - ac & cb + ab \\ ac + bc & bc + ac & - ab\end{vmatrix} \left[\text{ Taking out a, b and c common from the three columns }\right]\]

\[\begin{vmatrix}ab + bc + ca & ab + bc + ca & ab + bc + ca \\ ab + bc & - ac & cb + ab \\ ac + bc & bc + ac & - ab\end{vmatrix} \left[\text{ Applying }R_1 \text{ to }R_1 + R_2 + R_3 \right]\]

\[ = (ab + bc + ca)\begin{vmatrix}1 & 1 & 1 \\ ab + bc & - ac & cb + ab \\ ac + bc & bc + ac & - ab\end{vmatrix}\]

\[ = (ab + bc + ca)\begin{vmatrix}0 & 0 & 1 \\ 0 & - (ab + bc + ac) & cb + ab \\ ac + bc + ab & bc + ac + ab & - ab\end{vmatrix} \left[\text{ Applying }C_1 \text{ to }C_1 - C_3 \text{ and }C_2 \text{ to }C_2 - C_3 \right]\]

\[ = (ab + bc + ca)\begin{vmatrix}0 & - (ab + bc + ac) \\ ac + bc + ab & bc + ac + ab\end{vmatrix}\]

\[ = (ab + bc + ca)(ab + bc + ac )^2 \]

\[ = (ab + bc + ca )^3\]

Hence proved.