faktorisieren 8y^3-729x^6z^9

Lösung

faktorisieren

8 y^{3} - 729 x^{6} z^{9}

(2 y - 9 x^{2} z^{3}) (4 y^{2} + 18 y x^{2} z^{3} + 81 x^{4} z^{6})

Schritte zur Lösung

8 y^{3} - 729 x^{6} z^{9}

Schreibe

8 y^{3}

um:

(2 y)^{3}

Schreibe

729 x^{6} z^{9}

um:

(3^{2} x^{2} z^{3})^{3}

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

Wende Formel zur Differenz von dritten Potenzen an:

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

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

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

Vereinfache

(2 y - 3^{2} x^{2} z^{3}) ((2 y)^{2} + 3^{2} x^{2} z^{3} \cdot 2 y + (3^{2} x^{2} z^{3})^{2}) : (2 y - 9 x^{2} z^{3}) (4 y^{2} + 18 y x^{2} z^{3} + 81 x^{4} z^{6})

= (2 y - 9 x^{2} z^{3}) (4 y^{2} + 18 y x^{2} z^{3} + 81 x^{4} z^{6})

f a c t or 5^{2} - 7 x + 9

f a c t or 7 \cdot ia l (7 - 1)

f a c t or (\frac{π}{4}) + (\frac{16}{3})

f a c t or in (e^{6})

s im pl i f y (δ)^{'} + 10 δ (x)