faktorisieren b^6-64

Lösung

faktorisieren

b^{6} - 64

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

Schritte zur Lösung

b^{6} - 64

Wende Exponentenregel an

= (b^{2})^{3} - 4^{3}

Wende Formel zur Differenz von dritten Potenzen an:

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

(b^{2})^{3} - 4^{3} = (b^{2} - 4) (b^{4} + 4 b^{2} + 4^{2})

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

Faktorisiere

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

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

Faktorisiere

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

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

s im pl i f y i^{- 55}

\frac{1}{2} (20 x + 6) \geq x + 30

x^{2} + 5 x + 9 = 0

f a c t or 3 x^{3} - 6 x^{2} - 45 x

lo g_{4} (2 x + 5) \leq lo g_{4} (4 x - 3)