x^5=1 허근 곱 계산

$x ^{5} - 1 = ( x - 1 ) ( x ^{4} + x ^{3} + x ^{2} + x + 1 ) = 0$이므로 방정식$x ^{4} + x ^{3} + x ^{2} + x + 1 = 0$의 근이$\omega _{1} , \omega _{2} , \omega _{3} , \omega _{4}$이므로 $x ^{4} + x ^{3} + x ^{2} + x + 1 = ( x - \omega _{1} ) ( x - \omega _{2} ) ( x - \omega _{3} ) ( x - \omega _{4} )$이다. ⅰ) $x = 1$을 대입하면 $5 = ( 1 - \omega _{1} ) ( 1 - \omega _{2} ) ( 1 - \omega _{3} ) ( 1 - \omega _{4} )$ ⅱ) $x = - 1$을 대입하면 $1 = ( - 1 - \omega _{1} ) ( - 1 - \omega _{2} ) ( - 1 - \omega _{3} ) ( - 1 - \omega _{4} )$ $1 = ( 1 + \omega _{1} ) ( 1 + \omega _{2} ) ( 1 + \omega _{3} ) ( 1 + \omega _{4} )$ $\displaystyle \therefore \frac{3 {( 1 - \omega _{1} ) ( 1 - \omega _{2} ) ( 1 - \omega _{3} ) ( 1 - \omega _{4} )}}{( 1 + \omega _{1} ) ( 1 + \omega _{2} ) ( 1 + \omega _{3} ) ( 1 + \omega _{4} )} = 15$