Question

If $\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$,
where a,b,c are all different real no., then the determinant
$\left|\begin{array}{ccc}1& 1& 1\\ (x-a{)}^2& (x-b{)}^2& (x-c{)}^2\\ (x-b)(x-c)& (x-c)(x-a)& (x-a)(x-b)\end{array}\right|$
vanishes when

Answer 1

We have
$\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$
Taking $a,b,c$ from $C_1,C_2,C_3$ respectively
$\Rightarrow \mathrm{abc}\left|\begin{array}{ccc}\frac{1}{a}& \frac{1}{b}& \frac{1}{c}\\ 1& 1& 1\\ a^2& b^2& c^2\end{array}\right|$

$\Rightarrow \left|\begin{array}{ccc}bc& ac& ab\\ 1& 1& 1\\ a^2& b^2& c^2\end{array}\right|\Rightarrow -\left|\begin{array}{ccc}1& 1& 1\\ bc& ac& ab\\ a^2& b^2& c^2\end{array}\right|\Rightarrow \left|\begin{array}{ccc}1& 1& 1\\ a^2& b^2& c^2\\ bc& ac& ab\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$
From given determinant, we can write the value of required determinant by replacing
$a\to (x-a),b\to (x-b),c\to (x-c)$
Then
$D=\left|\begin{array}{ccc}1& 1& 1\\ (x-a{)}^2& (x-b{)}^2& (x-c{)}^2\\ (x-b)(x-c)& (x-c)(x-a)& (x-a)(x-b)\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c-3x)$
it vanishes when $3x=a+b+c$ or $a=b$ or $b=c$ or $c=a$