Question

If $1,{\alpha }_1,{\alpha }_2,{\alpha }_3$ and ${\alpha }_4$ be the roots of $x^5-1=0$, then $\frac{\omega -{\alpha }_1}{{\omega }^2-{\alpha }_1}.\frac{\omega -{\alpha }_2}{{\omega }^2-{\alpha }_2}.\frac{\omega -{\alpha }_3}{{\omega }^2-{\alpha }_3}.\frac{\omega -{\alpha }_4}{{\omega }^2-{\alpha }_4}=$

• A. $0$
• B. $\omega$
• C. ${\omega }^2$
• D. None of these

Answer 1

Answer: (B)

$\omega$

Since $1,{\alpha }_1,{\alpha }_2,{\alpha }_3,a_4$ are the roots of the equation $x^5-1=0.$

$\therefore (x^5-1)=(x-1)(x-{\alpha }_1)(x-{\alpha }_2)(x-{\alpha }_3)(x-{\alpha }_4)$

$\Rightarrow \frac{x^5-1}{x-1}=(x-{\alpha }_1)(x-{\alpha }_2)(x-{\alpha }_3)(x-{\alpha }_4)$ ...(1)

Putting $x=\omega$ in (1), we get

$\frac{{\omega }^5-1}{\omega -1}=(\omega -{\alpha }_1)(\omega -{\alpha }_2)(\omega -{\alpha }_3)(\omega -{\alpha }_4)$

$\Rightarrow \frac{\omega -1}{{\omega }^2-1}=(\omega -{\alpha }_1)(\omega -{\alpha }_2)(\omega -{\alpha }_3)(\omega -{\alpha }_4)$ ...(2)

and putting $x={\omega }^2$ in (1), we get

$\frac{{\omega }^{10}-1}{\omega -1}=({\omega }^2-{\alpha }_1)({\omega }^2-{\alpha }_2)({\omega }^2-{\alpha }_3)({\omega }^2-{\alpha }_4)$

$\Rightarrow \frac{\omega -1}{{\omega }^2-1}=({\omega }^2-{\alpha }_1)({\omega }^2-{\alpha }_2)({\omega }^2-{\alpha }_3)({\omega }^2-{\alpha }_4)$ ...(3)

Dividing (2) by (3), we get

$\frac{\omega -{\alpha }_1}{{\omega }^2-{\alpha }_1}.\frac{\omega -{\alpha }_2}{{\omega }^2-{\alpha }_2}.\frac{\omega -{\alpha }_3}{{\omega }^2-{\alpha }_3}.\frac{\omega -{\alpha }_4}{{\omega }^2-{\alpha }_4}=\frac{{({\omega }^2-1)}^2}{{(\omega -1)}^2}$

$=\frac{{\omega }^4+1-2{\omega }^2}{{\omega }^4+1-2\omega }=\frac{\omega +1-2{\omega }^2}{{\omega }^2+1-2\omega }$

$=\frac{-{\omega }^2-2{\omega }^2}{-\omega -2\omega }=\frac{-3{\omega }^2}{-3\omega }=\omega$