Question

If $1,d_1,d_2,d_3,d_4$ are roots of $x^5=1$ then the value of expression :$E=\frac{\omega -d_1}{{\omega }^2-d_1}\cdot \frac{\omega -d_2}{{\omega }^2-d_2}\cdot \frac{\omega -d_3}{{\omega }^2-d_3}\cdot \frac{\omega -d_4}{{\omega }^2-d_4}$ is
[Here $\omega$ is the cube root of unity]

• A. $\omega$
• B. ${\omega }^2$
• C. $1$
• D. $0$

Answer 1

The correct option is A $\omega$

We know, $x^5-1=(x-1)(x-d_1)(x-d_2)(x-d_3)(x-d_4)$
$\Rightarrow (x-d_1)(x-d_2)(x-d_3)(x-d_4)=\frac{x^5-1}{x-1}=1+x+x^2+x^3+x^4\cdots \left(1\right)$

Put $x=\omega$ in equation $\left(1\right)$

$\Rightarrow (\omega -d_1)(\omega -d_2)(\omega -d_3)(\omega -d_4)=1+\omega +{\omega }^2+{\omega }^3+{\omega }^4=1+\omega$

$\{\because 1+\omega +{\omega }^2=0,{\omega }^3=1\}$

Put $x={\omega }^2$ in equation (1)

$\Rightarrow ({\omega }^2-d_1)({\omega }^2-d_2)({\omega }^2-d_3)({\omega }^2-d_4)=1+{\omega }^2+{\omega }^4+{\omega }^6+{\omega }^8=1+{\omega }^2$

$\therefore E=\frac{\omega -d_1}{{\omega }^2-d_1}\cdot \frac{\omega -d_2}{{\omega }^2-d_2}\cdot \frac{\omega -d_3}{{\omega }^2-d_3}\cdot \frac{\omega -d_4}{{\omega }^2-d_4}=\frac{1+\omega }{1+{\omega }^2}$

$E=\frac{\omega (\omega +1)}{\omega (1+{\omega }^2)}=\frac{\omega (\omega +1)}{1+\omega }=\omega$