faktorisieren x^9+y^9

Lösung

faktorisieren

x^{9} + y^{9}

(x + y) (x^{2} - x y + y^{2}) (x^{6} - x^{3} y^{3} + y^{6})

Schritte zur Lösung

x^{9} + y^{9}

Wende Exponentenregel an

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

Wende Formel zur Summe von dritten Potenzen an:

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

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

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

Vereinfache

(x^{3})^{2} - x^{3} y^{3} + (y^{3})^{2} : x^{6} - x^{3} y^{3} + y^{6}

= (x^{3} + y^{3}) (x^{6} - x^{3} y^{3} + y^{6})

Faktorisiere

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

= (x + y) (x^{2} - x y + y^{2}) (x^{6} - x^{3} y^{3} + y^{6})

s im pl i f y (\frac{y}{3})^{- 7}

f a c t or 24 x y + 25 y^{2} - 20 x - 30 y^{3}

\frac{4 x + 6}{x ^{2} + 5 x + 6} \geq 1

s im pl i f y \frac{4}{3} - \frac{2}{4}

9 y - 6 + y + 8 = 42