Question

If $ \begin{vmatrix}1&1&1\\ a&b&c\\ a^{3}&b^{3}&c^{3}\end{vmatrix}$ $=(a-b)(b-c)(c-a)(a+b+c),$ where $a, b, c$ are all different, then the determinant
$ \begin{vmatrix}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{vmatrix}$
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, $ \begin{vmatrix}1&1&1\\ a&b&c\\ a^{3}&b^{3}&c^{3}\end{vmatrix}$
$=(a-b)(b-c)(c-a)(a+b+c)$
Then $ \Delta=\begin{vmatrix}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{vmatrix}$
$=(a-b)(b-c)(c-a)(3 x-a-b-c)$
Now given that $a, b, c$ are all different then $D=0$
$\Rightarrow x=\frac{1}{3}(a+b+c)$