faktorisieren x^6-3x^4+3x^2-1

Lösung

faktorisieren

x^{6} - 3 x^{4} + 3 x^{2} - 1

(x + 1)^{3} (x - 1)^{3}

Schritte zur Lösung

x^{6} - 3 x^{4} + 3 x^{2} - 1

Angenommen

u = x^{2}

= u^{3} - 3 u^{2} + 3 u - 1

Faktorisiere

u^{3} - 3 u^{2} + 3 u - 1 : (u - 1) (u - 1) (u - 1)

= (u - 1) (u - 1) (u - 1)

Setze in

u = x^{2}

ein

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

Faktorisiere

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

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

Faktorisiere

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

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

Faktorisiere

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

= (x + 1) (x - 1) (x + 1) (x - 1) (x + 1) (x - 1)

Fasse zusammen

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

f a c t or e^{a t} u (- t)

f a c t or 2 cos^{2} (x) - 3 cos (x) + 1

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

f a c t or 1

f a c t or 2 (x + y)^{2} + 9 (x + y) - 5