Question

State whether the following statement is true or false.

$\left|\begin{array}{ccc}{(b+c)}^2& a^2& a^2\\ b^2& {(c+a)}^2& b^2\\ c^2& c^2& (a+b{)}^2\end{array}\right|=2abc(a+b+c{)}^3$

• A. True
• B. False

Answer 1

The correct option is A True

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

$\Rightarrow c_2\to c_2-c_1$

$c_3\to c_3-c_1$

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

$\Rightarrow$ Taking common $(a+b+c)$ from $c_2$ & $c_3$

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

$\Rightarrow R_1\to R_1-(R_2+R_3)$

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

$\Rightarrow c_2\to (c_2+\frac{1}{b}{c}_1)$

$c_3\to (c_3+\frac{1}{c}{c}_1)$

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

Expanding along $R_1$

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

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

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

Hence the statement is true.