faktorisieren 3a^{12}+1+3a^6+a^{18}

Lösung

faktorisieren

3 a^{12} + 1 + 3 a^{6} + a^{18}

(a^{2} + 1)^{3} (a^{12} - 3 a^{10} + 6 a^{8} - 7 a^{6} + 6 a^{4} - 3 a^{2} + 1)

Schritte zur Lösung

3 a^{12} + 1 + 3 a^{6} + a^{18}

Schreibe in der Standard Form

= a^{18} + 3 a^{12} + 3 a^{6} + 1

Angenommen

u = a^{2}

= u^{9} + 3 u^{6} + 3 u^{3} + 1

Faktorisiere

u^{9} + 3 u^{6} + 3 u^{3} + 1 : (u + 1) (u + 1) (u + 1) (u^{6} - 3 u^{5} + 6 u^{4} - 7 u^{3} + 6 u^{2} - 3 u + 1)

= (u + 1) (u + 1) (u + 1) (u^{6} - 3 u^{5} + 6 u^{4} - 7 u^{3} + 6 u^{2} - 3 u + 1)

Setze in

u = a^{2}

ein

= (a^{2} + 1) (a^{2} + 1) (a^{2} + 1) (a^{12} - 3 a^{10} + 6 a^{8} - 7 a^{6} + 6 a^{4} - 3 a^{2} + 1)

Wende Exponentenregel an:

aaa = a^{3}

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

= (a^{2} + 1)^{3} (a^{12} - 3 a^{10} + 6 a^{8} - 7 a^{6} + 6 a^{4} - 3 a^{2} + 1)

e x p an d (6 + x)^{3}

e x p an d (a + b)^{3} + 1

s im pl i f y x (x + y)^{3} - 3 x^{2} y (x + y)

f a c t or 9 x^{2} + 12 x + 4 - 16 y^{2}

s im pl i f y 2 sin^{2} (a) + 2 cos^{2} (a)