Question

If $\alpha ,\beta$ are the roots of $x^2+x+1=0$, $\gamma ,\delta$ are roots of $x^2+x+q=0$ then $q-\frac{(\gamma -\alpha )(\delta -\alpha )(\beta -\gamma )(\beta -\delta )}{q-1}$ is equal to, $(q\ne 1)$......

Answer 1

We know that $\alpha$ and $\beta$ are roots of $x^2+x+1=0$.

So, $\alpha +\beta =-1$

Substitute value of $x=\alpha$ and $x=\beta$ in the given equation.

${\alpha }^2+\alpha +1={\beta }^2+\beta +1=0$ .....(1)

and similarly

$\gamma +\delta =-1$ and $\gamma \delta =q$ ....(2)

Simplifying the given expression:

$q-\frac{(\gamma -\alpha )(\delta -\alpha )(\beta -\gamma )(\beta -\delta )}{q-1}$

$=q-\frac{(\gamma \delta -\alpha (\gamma +\delta )+{\alpha }^2)(\gamma \delta -\beta (\gamma +\delta )+{\beta }^2)}{q-1}$

Using (1) and (2) to simplify, we get:

$=q-\frac{(q+\alpha +{\alpha }^2)(q+\beta +{\beta }^2)}{q-1}$

$=q-\frac{(q-1)(q-1)}{q-1}$

$=q-(q-1)$

$=1$