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, then the determinant
$\left|\begin{array}{ccc}1& 1& 1\\ {\left(x-a\right)}^2& {\left(bx-b\right)}^2& {\left(x-c\right)}^2\\ \left(x-b\right)\left(x-c\right)& \left(x-c\right)\left(x-a\right)& \left(x-a\right)\left(x-b\right)\end{array}\right|$
vanishes when

• A. $a+b+c=0$
• B. $x=\frac{1}{3}(a+b+c)$
• C. $x=\frac{1}{2}(a+b+c)$
• D. $x=a+b+c$

Answer 1

Answer: (B)

We have, $\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|$
$=(a-b)(b-c)(c-a)(a+b+c)$
Then $\Delta =\left|\begin{array}{ccc}1& 1& 1\\ {\left(x-a\right)}^2& {\left(x-b\right)}^2& {\left(x-c\right)}^2\\ \left(x-b\right)\left(x-c\right)& \left(x-c\right)\left(x-a\right)& \left(x-a\right)\left(x-b\right)\end{array}\right|$
$=(a-b)(b-c)(c-a)(3x-a-b-c)$
Now given that $a,b,c$ are all different then $D=0$
$\Rightarrow x=\frac{1}{3}(a+b+c)$