Question

Using properties of deteminants, prove that $\left|\begin{array}{ccc}\frac{{(a+b)}^2}{c}& c& c\\ a& \frac{{(b+c)}^2}{a}& a\\ b& b& \frac{{(c+a)}^2}{b}\end{array}\right|=2{(a+b+c)}^3$.

Answer 1

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

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

On applying $C_2\to C_2-C_{1}and{C}_3\to C_3-C_1$, we get-

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

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

Taking $(a+b+c)$ common from $C_{2}and{C}_3$,we get-

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

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

On applying $R_1\to R_1-(R_2+R_3)$, we get-

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

On applying $C_3\to C_3+\frac{1}{b}{C}_1$, we get-

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

Expanding the determinant now-

$\frac{(a+b+c{)}^2}{abc}\left[2ab\right(bc+ab+c^2+ac-ac-a^2)+2b(a^{2}c+a^3-a^{2}b\left)\right]$

On simplifying, we get-

$\frac{(a+b+c{)}^2}{abc}\times 2abc(a+b+c)$

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

Hence,proved.