faktorisieren x^{12}-64

Lösung

faktorisieren

x^{12} - 64

(x^{2} + 2) (x^{2} - 2) (x^{4} + 2 x^{2} + 4) (x^{4} - 2 x^{2} + 4)

Schritte zur Lösung

x^{12} - 64

Wende Exponentenregel an

= (x^{4})^{3} - 4^{3}

Wende Formel zur Differenz von dritten Potenzen an:

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

(x^{4})^{3} - 4^{3} = (x^{4} - 4) (x^{8} + 4 x^{4} + 4^{2})

= (x^{4} - 4) (x^{8} + 4 x^{4} + 4^{2})

Faktorisiere

x^{4} - 4 : (x^{2} + 2) (x^{2} - 2)

= (x^{2} + 2) (x^{2} - 2) (x^{8} + 4 x^{4} + 4^{2})

Faktorisiere

x^{8} + 4 x^{4} + 4^{2} : (x^{4} + 4 + 2 x^{2}) (x^{4} + 4 - 2 x^{2})

= (x^{2} + 2) (x^{2} - 2) (x^{4} + 4 + 2 x^{2}) (x^{4} + 4 - 2 x^{2})

= (x^{2} + 2) (x^{2} - 2) (x^{4} + 2 x^{2} + 4) (x^{4} - 2 x^{2} + 4)

\frac{9 x}{2} - \frac{x}{6} = 2

y (y - 1) - (y - 5)^{2} = 2

\frac{1}{3} (\frac{x - 1}{2}) < x - 3

\frac{3 x + 7}{2} = - 4

\frac{3}{2} x - 1 < 1 - 3 x < \frac{4}{3} - 5 x