Complex Roots

Since $\alpha ,\beta ,\gamma ,\delta$ are the roots of the equation $x^{4} + x^{3} + x^{2} + x + 1 = 0$ ;

$\therefore x^{4}+x^{3}+x^{2}+x+1=(x-\alpha )(x-\beta )(x-\gamma )(x-\delta )$

Putting $x=1$ ;

$\therefore 1+1+1+1+1=(1-\alpha)(1-\beta)(1-\gamma)(1-\delta)$

$\therefore (1-\alpha)(1-\beta)(1-\gamma)(1-\delta)=\boxed5$

Whoah, amazingly, it took me less than 30 seconds!

If a, b, c, and d are the roots of the above equation, then $1-a, 1-b, 1-c, 1-d$ are the roots of $(1-x)^4 +(1-x)^3+(1-x)^2+(1-x)+1$ . A quick look tells us that when expanded, the last term of this is 5, and by vieta's formulas, this is the product of all the roots of the equation, namely $1-a, 1-b, 1-c, 1-d$