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. $1$
• B. $\omega$
• C. ${\omega }^2$
• D. None of these

Answer 1

The correct option is B $\omega$

Since $1,{\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4$ are roots of the equation $x^5-1=0$

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

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

Putting $x=w$ in 1 we get,

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

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

Putting $x=w^2$ in 1 we get

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

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

Dividing 2 by 3 we get,

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

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