faktorisieren r^9-8(r^3-1)^3

Lösung

faktorisieren

r^{9} - 8 (r^{3} - 1)^{3}

- (r^{3} - 2) (7 r^{6} - 10 r^{3} + 4)

Schritte zur Lösung

r^{9} - 8 (r^{3} - 1)^{3}

Schreibe

r^{9} - 8 (r^{3} - 1)^{3}

um:

(r^{3})^{3} - (2 (r^{3} - 1))^{3}

= (r^{3})^{3} - (2 (r^{3} - 1))^{3}

Wende Formel zur Differenz von dritten Potenzen an:

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

(r^{3})^{3} - (2 (r^{3} - 1))^{3} = (r^{3} - 2 (r^{3} - 1)) (r^{6} + 2 r^{3} (r^{3} - 1) + 2^{2} (r^{3} - 1)^{2})

= (r^{3} - 2 (r^{3} - 1)) (r^{6} + 2 r^{3} (r^{3} - 1) + 2^{2} (r^{3} - 1)^{2})

Fasse zusammen

= (r^{3} - 2 (r^{3} - 1)) (r^{6} + 2 r^{3} (r^{3} - 1) + 4 (r^{3} - 1)^{2})

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

= (r^{3} - 2 (r - 1) (r^{2} + r + 1)) (r^{6} + 2 r^{3} (r^{3} - 1) + 4 (r^{3} - 1)^{2})

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

4 (r^{3} - 1)^{2} = 4 (r - 1)^{2} (r^{2} + r + 1)^{2}

= (r^{3} - 2 (r - 1) (r^{2} + r + 1)) (r^{6} + 2 r^{3} (r - 1) (r^{2} + r + 1) + 4 (r - 1)^{2} (r^{2} + r + 1)^{2})

Faktorisiere

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

= - (r^{3} - 2) (r^{6} + 2 r^{3} (r - 1) (r^{2} + r + 1) + 4 (r - 1)^{2} (r^{2} + r + 1)^{2})

Multipliziere aus

r^{6} + 2 r^{3} (r - 1) (r^{2} + r + 1) + 4 (r - 1)^{2} (r^{2} + r + 1)^{2} : 7 r^{6} - 10 r^{3} + 4

= - (r^{3} - 2) (7 r^{6} - 10 r^{3} + 4)

2 x^{2} + 4 = 22

s im pl i f y \frac{15}{2}

5 x + 3 < 18

6 x^{2} - 19 x + 10 = 0

s im pl i f y - 70