Question

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

• A. ${\omega }^2$
• B. $1$
• C. $\omega$
• D. $(\omega -{\alpha }_1)(\omega -{\alpha }_2)(\omega -{\alpha }_3)(\omega -{\alpha }_4)$

Answer 1

The correct option is C $\omega$

$x^5-1=0$ has roots $1,{\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4$
$\therefore (x^5-1)=(x-1)(x-{\alpha }_1)(x-{\alpha }_2)(x-{\alpha }_3)(x-{\alpha }_4)$
$\Rightarrow \frac{x^5-1}{x-1}=(x-{\alpha }_1)(x-{\alpha }_2)(x-{\alpha }_3)(x-{\alpha }_4)$ ........ (1)
Putting $x=\omega$ (1) we have
$\frac{{\omega }^5-1}{\omega -1}=(\omega -{\alpha }_1)(\omega -{\alpha }_2)(\omega -{\alpha }_3)(\omega -{\alpha }_4)$
$\frac{{\omega }^2-1}{\omega -1}=(\omega -{\alpha }_1)(\omega -{\alpha }_2)(\omega -{\alpha }_3)(\omega -{\alpha }_4)$ ....... (2)
and putting $x={\omega }^2$ in (1) we have
$\frac{{\omega }^{10}-1}{{\omega }^2-1}=({\omega }^2-{\alpha }_1)({\omega }^2-{\alpha }_2)({\omega }^2-{\alpha }_3)({\omega }^2-{\alpha }_4)$
$\Rightarrow \frac{\omega -1}{{\omega }^2-1}=({\omega }^2-{\alpha }_1)({\omega }^2-{\alpha }_2)({\omega }^2-{\alpha }_3)({\omega }^2-{\alpha }_4)$ ....... (3)
Dividing (2) by (3)
then $\frac{\omega -{\alpha }_1}{{\omega }^2-{\alpha }_1}\cdot \frac{\omega -{\alpha }_2}{{\omega }^2-{\alpha }_2}\cdot \frac{\omega -{\alpha }_3}{{\omega }^2-{\alpha }_3}\cdot \frac{\omega -{\alpha }_4}{{\omega }^2-{\alpha }_4}\cdot =\frac{({\omega }^2-1{)}^2}{(\omega -1{)}^2}$
$=\frac{{\omega }^4+1-2{\omega }^2}{{\omega }^2+1-2\omega }$
$=\frac{\omega +1-2{\omega }^2}{{\omega }^2+1-2\omega }$
$=\frac{-{\omega }^2-2{\omega }^2}{-\omega -2\omega }$
$=\frac{-3{\omega }^2}{-3\omega }$
$=\omega$
Ans: C