Question

cum descompun in factori ireductibili polinomul f=x^6-1 apartine C[x].

Answer 1

x^6 - 1 = (x^3-1)(x^3+1)=(x-1)(x^2+x+1)(x+1)(x^2-x+1)

Answer 2

$x^6-1=(x^3-1)(x^3+1)=(x-1)(x^2+x+1)(x+1)(x^2-x+1)=$
$(x-1)(x+1)(x-\dfrac{1+i\sqrt3}{2})(x-\dfrac{1-i\sqrt3}{2})(x+\dfrac{1+i\sqrt3}{2})(x+\dfrac{1-i\sqrt3}{2})$
Am folosit faptul ca solutiile ecuatiei
$x^2-x+1=0\ sunt\ x_{1,2}=\dfrac{1\pm i\sqrt3}{2}$, iar ale ecuatiei
$x^2+x+1=0\ sunt\ x_{1,2}=\dfrac{-1\pm i\sqrt3}{2}$