Question

If $\left|\begin{array}{ccc}x+a& a^2& a^3\\ x+b& b^2& b^3\\ x+c& c^2& c^3\end{array}\right|=0$ and $a\ne b\ne c,$ then the value of $x$ is:

• A. $\frac{-abc}{ab+bc+ca}$
• B. $\frac{abc}{(a-b)(b-c)(c-a)}$
• C. $\frac{abc}{ab+bc+ca}$
• D. $\frac{(a-b)(b-c)(c-a)}{abc}$

Answer 1

The correct option is A $\frac{-abc}{ab+bc+ca}$

Given: $\Delta =\left|\begin{array}{ccc}x+a& a^2& a^3\\ x+b& b^2& b^3\\ x+c& c^2& c^3\end{array}\right|=0$
The given determinant can be expressed as a sum of two determinants as:
$\left|\begin{array}{ccc}x& a^2& a^3\\ x& b^2& b^3\\ x& c^2& c^3\end{array}\right|+\left|\begin{array}{ccc}a& a^2& a^3\\ b& b^2& b^3\\ c& c^2& c^3\end{array}\right|=0$
$\Rightarrow x\left|\begin{array}{ccc}1& a^2& a^3\\ 1& b^2& b^3\\ 1& c^2& c^3\end{array}\right|+abc\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=0$
$\Rightarrow x(a-b)(b-c)(c-a)(ab+bc+ca)+abc(a-b)(b-c)(c-a)=0$
$\Rightarrow x=\frac{-abc}{ab+bc+ca}$