faktorisieren (45b^7)-3

Lösung

faktorisieren

(45 b^{7}) - 3

3 (715 b - 1) (1 5^{\frac{6}{7}} b^{6} + 1 5^{\frac{5}{7}} b^{5} + 1 5^{\frac{4}{7}} b^{4} + 1 5^{\frac{3}{7}} b^{3} + 1 5^{\frac{2}{7}} b^{2} + 715 b + 1)

Schritte zur Lösung

(45 b^{7}) - 3

Entferne die Klammern:

(a) = a

= 45 b^{7} - 3

Klammere gleiche Terme aus

3 : 3 (15 b^{7} - 1)

= 3 (15 b^{7} - 1)

Faktorisiere

15 b^{7} - 1 : (715 b - 1) ((715)^{6} b^{6} + (715)^{5} b^{5} + (715)^{4} b^{4} + (715)^{3} b^{3} + (715)^{2} b^{2} + 715 b + 1)

= 3 (715 b - 1) ((715)^{6} b^{6} + (715)^{5} b^{5} + (715)^{4} b^{4} + (715)^{3} b^{3} + (715)^{2} b^{2} + 715 b + 1)

Fasse zusammen

= 3 (715 b - 1) (1 5^{\frac{6}{7}} b^{6} + 1 5^{\frac{5}{7}} b^{5} + 1 5^{\frac{4}{7}} b^{4} + 1 5^{\frac{3}{7}} b^{3} + 1 5^{\frac{2}{7}} b^{2} + 715 b + 1)

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