Question

Prove that $\left|\begin{array}{ccc}1+a^2+a^4& 1+ab+a^2{b}^2& 1+ac+a^2{c}^2\\ 1+ab+a^2{b}^2& 1+b^2+b^4& 1+bc+b^2{c}^2\\ 1+ac+a^2{c}^2& 1+bc+b^2{c}^2& 1+c^2+c^4\end{array}\right|={(a-b)}^2{(b-c)}^2{(c-a)}^2$

Answer 1

$\left|\begin{array}{ccc}1+a^2+a^4& 1+ab+a^2{b}^2& 1+ac+a^2{c}^2\\ 1+ab+a^2{b}^2& 1+b^2+b^4& 1+bc+b^2{c}^2\\ 1+ac+a^2{c}^2& 1+bc+b^2{c}^2& 1+c^2+c^4\end{array}\right|$

$=(a-b{)}^2(b-c{)}^2(c-a{)}^2$

L.H.S.

If $(a-b{)}^2$ is a factor of the determinant then it we put $(a-b{)}^2=0$

$\Rightarrow C=0$

put

so $a=b$

$\left|\begin{array}{ccc}1+a^2+a^4& 1+a^2+a^4& 1+ac+a^2{c}^2\\ 1+a^2+a^4& 1+a^2+a^4& 1+ac+a^2{c}^2\\ 1+ac+a^2{c}^2& 1+ac+b^2{c}^2& 1+c^2+c^4\end{array}\right|$$=0$

Similarly

$(b-c{)}^2$ & $(c-a{)}^2$ also the factor of determinants

Thus $(a-b{)}^2(b-c{)}^2(c-a{)}^2$ is a factor of determinant

Thus $\Delta =(a-b{)}^2(b-c{)}^2(c-a{)}^2$.