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

Lösung

faktorisieren

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

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

Schritte zur Lösung

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

Schreibe

x^{3} y^{6}

um:

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

Schreibe

8 z^{9}

um:

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

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

Wende Formel zur Differenz von dritten Potenzen an:

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

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

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

Vereinfache

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

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

f (y) = y^{(4)}

\frac{d}{d y} ((x) (x^{2}))

s im pl i f y β (1.1)

f a c t or x + y + 3 x z + 3 yz

f a c t or y^{12} - 8