(d^3+1)^3(d^2+d+1)=0

Lösung

(d^{3} + 1)^{3} (d^{2} + d + 1) = 0

d = - 1, d = \frac{1}{2} + i \frac{3}{2}, d = \frac{1}{2} - i \frac{3}{2}, d = - \frac{1}{2} + i \frac{3}{2}, d = - \frac{1}{2} - i \frac{3}{2}

Schritte zur Lösung

(d^{3} + 1)^{3} (d^{2} + d + 1) = 0

Faktorisiere

(d^{3} + 1)^{3} (d^{2} + d + 1) : (d + 1)^{3} (d^{2} - d + 1)^{3} (d^{2} + d + 1)

(d + 1)^{3} (d^{2} - d + 1)^{3} (d^{2} + d + 1) = 0

Anwendung des Nullfaktorprinzips: Wenn

ab = 0

dann

a = 0

oder

b = 0

d + 1 = 0 or d^{2} - d + 1 = 0 or d^{2} + d + 1 = 0

Löse

d + 1 = 0 : d = - 1

Löse

d^{2} - d + 1 = 0 : d = \frac{1}{2} + i \frac{3}{2}, d = \frac{1}{2} - i \frac{3}{2}

Löse

d^{2} + d + 1 = 0 : d = - \frac{1}{2} + i \frac{3}{2}, d = - \frac{1}{2} - i \frac{3}{2}

Die Lösungen sind

d = - 1, d = \frac{1}{2} + i \frac{3}{2}, d = \frac{1}{2} - i \frac{3}{2}, d = - \frac{1}{2} + i \frac{3}{2}, d = - \frac{1}{2} - i \frac{3}{2}

x^{6} + 11 x^{4} + 19 x^{2} - 91 = 0

x^{2} (x^{2} - 10) + 9 = (x + 1) \cdot (x - 1)

x^{4} - 20 x^{2} + 36 = 0

3 + 3 x^{3} = 9

(x + 1) (x^{2} - 2) = (x + 3) 3