faktorisieren 8(a+1)^3-1

Lösung

faktorisieren

8 (a + 1)^{3} - 1

(2 a + 1) (4 a^{2} + 10 a + 7)

Schritte zur Lösung

8 (a + 1)^{3} - 1

Schreibe

8 (a + 1)^{3} - 1

um:

(2 (a + 1))^{3} - 1^{3}

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

Wende Formel zur Differenz von dritten Potenzen an:

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

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

= (2 (a + 1) - 1) (2^{2} (a + 1)^{2} + 2 (a + 1) + 1)

Fasse zusammen

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

Multipliziere aus

2 (a + 1) - 1 : 2 a + 1

= (2 a + 1) (4 (a + 1)^{2} + 2 (a + 1) + 1)

Multipliziere aus

4 (a + 1)^{2} + 2 (a + 1) + 1 : 4 a^{2} + 10 a + 7

= (2 a + 1) (4 a^{2} + 10 a + 7)

f a c t or 2 x^{2} + (- 8 x)

s im pl i f y [0.3]

f a c t or 2 sin^{2} (x) + sin (x) - 1

f (x) = (x)^{5}

s im pl i f y [- 9.8]