faktorisieren 64x^6-1

Lösung

faktorisieren

64 x^{6} - 1

(2 x + 1) (2 x - 1) (4 x^{2} + 2 x + 1) (4 x^{2} - 2 x + 1)

Schritte zur Lösung

64 x^{6} - 1

Wende Exponentenregel an

= (4 x^{2})^{3} - 1^{3}

Wende Formel zur Differenz von dritten Potenzen an:

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

(4 x^{2})^{3} - 1^{3} = (4 x^{2} - 1) (4^{2} x^{4} + 4 x^{2} + 1)

= (4 x^{2} - 1) (4^{2} x^{4} + 4 x^{2} + 1)

Faktorisiere

4 x^{2} - 1 : (2 x + 1) (2 x - 1)

= (2 x + 1) (2 x - 1) (4^{2} x^{4} + 4 x^{2} + 1)

Faktorisiere

4^{2} x^{4} + 4 x^{2} + 1 : (4 x^{2} + 1 + 2 x) (4 x^{2} + 1 - 2 x)

= (2 x + 1) (2 x - 1) (4 x^{2} + 1 + 2 x) (4 x^{2} + 1 - 2 x)

= (2 x + 1) (2 x - 1) (4 x^{2} + 2 x + 1) (4 x^{2} - 2 x + 1)

f a c t or 36 x^{2} - 121 y^{2}

2 d + 24 + 3 d = 84

(x - 4) (x + 8) < 0

a^{2} - 8 a + 12 < 0

- 16 x^{2} + 64 x + 80 = 0