Question

$\left|\begin{array}{ccc}{(b+c)}^2& a^2& a^2\\ b^2& {(c+a)}^2& b^2\\ c^2& c^2& {(a+b)}^2\end{array}\right|$ is equal to

• A. $abc{(a+b+c)}^2$
• B. $2abc{(a+b+c)}^2$
• C. $2abc{(a+b+c)}^3$
• D. $2abc(a+b+c)$

Answer 1

The correct option is C $2abc{(a+b+c)}^3$

$\Delta =\left|\begin{array}{ccc}{(b+c)}^2& a^2& a^2\\ b^2& {(c+a)}^2& b^2\\ c^2& c^2& {(a+b)}^2\end{array}\right|$

Applying $C_1\to C_1-C_3;C_2\to C_2-C_3,$ we get

$\Delta =\left|\begin{array}{ccc}{(b+c)}^2-a^2& 0& a^2\\ 0& {(c+a)}^2-b^2& b^2\\ c^2-{(a+b)}^2& c^2-{(a+b)}^2& {(a+b)}^2\end{array}\right|$

$\Delta =\left|\begin{array}{ccc}(b+c+a)(b+c-a)& 0& a^2\\ 0& (c+a+b)(c+a-b)& b^2\\ (c+a+b)(c-a-b)& (c+a+b)(c-a-b)& {(a+b)}^2\end{array}\right|$

$={(a+b+c)}^2\left|\begin{array}{ccc}b+c-a& 0& a^2\\ 0& c+a-b& b^2\\ c-a-b& c-a-b& {(a+b)}^2\end{array}\right|$

Applying $R_3\to R_3-(R_1+R_2),$

we get

$\Delta ={(a+b+c)}^2\left|\begin{array}{ccc}b+c-a& 0& a^2\\ 0& c+a-b& b^2\\ -2b& -2a& 2ab\end{array}\right|$

$=\frac{{(a+b+c)}^2}{ab}\left|\begin{array}{ccc}ab+ac& a^2& a^2\\ b^2& bc+ba& b^2\\ 0& 0& 2ab\end{array}\right|$

$=\frac{{(a+b+c)}^2}{ab}.ab.2ab\left|\begin{array}{ccc}b+c& a& a\\ b& c+a& b\\ 0& 0& 1\end{array}\right|$

$=2ab{(a+b+c)}^2\left|\begin{array}{cc}b+c& a\\ b& c+a\end{array}\right|$

$=2ab{(a+b+c)}^2(a+b+c)=2ab{(a+b+c)}^3$