faktorisieren 1-128x^7y^7

Lösung

faktorisieren

1 - 128 x^{7} y^{7}

- (2 x y - 1) (64 x^{6} y^{6} + 32 x^{5} y^{5} + 16 x^{4} y^{4} + 8 x^{3} y^{3} + 4 x^{2} y^{2} + 2 x y + 1)

Schritte zur Lösung

1 - 128 x^{7} y^{7}

Schreibe

128 x^{7} y^{7}

um:

(2 x y)^{7}

= 1 - (2 x y)^{7}

Klammere gleiche Terme aus

- 1

= - ((2 x y)^{7} - 1)

Faktorisiere

(2 x y)^{7} - 1 : (2 x y - 1) ((2 x y)^{6} + (2 x y)^{5} + (2 x y)^{4} + (2 x y)^{3} + (2 x y)^{2} + 2 x y + 1)

= - (2 x y - 1) ((2 x y)^{6} + (2 x y)^{5} + (2 x y)^{4} + (2 x y)^{3} + (2 x y)^{2} + 2 x y + 1)

Vereinfache

- (2 x y - 1) ((2 x y)^{6} + (2 x y)^{5} + (2 x y)^{4} + (2 x y)^{3} + (2 x y)^{2} + 2 x y + 1) : - (2 x y - 1) (64 x^{6} y^{6} + 32 x^{5} y^{5} + 16 x^{4} y^{4} + 8 x^{3} y^{3} + 4 x^{2} y^{2} + 2 x y + 1)

= - (2 x y - 1) (64 x^{6} y^{6} + 32 x^{5} y^{5} + 16 x^{4} y^{4} + 8 x^{3} y^{3} + 4 x^{2} y^{2} + 2 x y + 1)

s im pl i f y [1.3]

f a c t or \frac{x ^{3} + y ^{3}}{x + y}

f a c t or a x + a y + a z + x + y + z

f a c t or x + 2 2 x y + 2 y

f a c t or (123) - 1