faktorisieren a^9+b^9

Lösung

faktorisieren

a^{9} + b^{9}

(a + b) (a^{2} - ab + b^{2}) (a^{6} - a^{3} b^{3} + b^{6})

Schritte zur Lösung

a^{9} + b^{9}

Wende Exponentenregel an

= (a^{3})^{3} + (b^{3})^{3}

Wende Formel zur Summe von dritten Potenzen an:

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

(a^{3})^{3} + (b^{3})^{3} = (a^{3} + b^{3}) ((a^{3})^{2} - a^{3} b^{3} + (b^{3})^{2})

= (a^{3} + b^{3}) ((a^{3})^{2} - a^{3} b^{3} + (b^{3})^{2})

Vereinfache

(a^{3})^{2} - a^{3} b^{3} + (b^{3})^{2} : a^{6} - a^{3} b^{3} + b^{6}

= (a^{3} + b^{3}) (a^{6} - a^{3} b^{3} + b^{6})

Faktorisiere

a^{3} + b^{3} : (a + b) (a^{2} - ab + b^{2})

= (a + b) (a^{2} - ab + b^{2}) (a^{6} - a^{3} b^{3} + b^{6})

2 x + 1 = 2 x - 1

f a c t or 10 x^{2} - 9 x - 9

3^{2} + 2^{2} = x^{2}

6 (u + 1) - 10 = 4 (u - 1) + 2 u

f a c t or 3 x^{2} - 19 x - 14