Question

If a, b, c be non-zero real numbers such that $\left|\begin{array}{ccc}bc& ca& ab\\ ca& ab& bc\\ ab& bc& ca\end{array}\right|=0$ where $\omega$ be an imaginary cube root of unity, then

• A. $\frac{1}{a}+\frac{1}{b\omega }+\frac{1}{c{\omega }^2}=0$
• B. $\frac{1}{b}+\frac{1}{c\omega }+\frac{1}{a{\omega }^2}=0$
• C. $\frac{1}{c}+\frac{1}{a\omega }+\frac{1}{b{\omega }^2}=0$
• D. $a+b+c=0$

Answer 1

Answer: (A), (B), (C)

$\frac{1}{a}+\frac{1}{b\omega }+\frac{1}{c{\omega }^2}=0$
$\frac{1}{b}+\frac{1}{c\omega }+\frac{1}{a{\omega }^2}=0$
$\frac{1}{c}+\frac{1}{a\omega }+\frac{1}{b{\omega }^2}=0$

Let $ab=x$, $bc=y$, $ca=z$
$D=\left|\begin{array}{ccc}y& z& x\\ z& x& y\\ x& y& z\end{array}\right|=x^3+y^3+z^3-3xyz$
$\therefore x^3+y^3+z^3-3xyz=0$
$(x+y{\omega }^2+z\omega )(x\omega +y+z{\omega }^2)(x{\omega }^2+y\omega +z)=0$
$x+y{\omega }^2+z\omega =0.........\left(1\right)$
$x\omega +y+z{\omega }^2=0.........\left(2\right)$
$x{\omega }^2+y\omega +z=0.........\left(3\right)$
From (1), $ab+bc{\omega }^2+ca\omega =0$
Divide by abc $(i.e.,abc{\omega }^3)$
$\frac{1}{c}+\frac{1}{a\omega }+\frac{1}{b{\omega }^2}=0$
Hence, option C is correct.
Similarly, from (2) and (3), options (A) and (B) are correct.