faktorisieren 27x^6y^{12}-8z^9

Lösung

faktorisieren

27 x^{6} y^{12} - 8 z^{9}

(3 x^{2} y^{4} - 2 z^{3}) (9 x^{4} y^{8} + 6 x^{2} y^{4} z^{3} + 4 z^{6})

Schritte zur Lösung

27 x^{6} y^{12} - 8 z^{9}

Schreibe

27 x^{6} y^{12}

um:

(3 x^{2} y^{4})^{3}

Schreibe

8 z^{9}

um:

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

= (3 x^{2} y^{4})^{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})

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

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

Vereinfache

(3 x^{2} y^{4} - 2 z^{3}) ((3 x^{2} y^{4})^{2} + 3 x^{2} y^{4} \cdot 2 z^{3} + (2 z^{3})^{2}) : (3 x^{2} y^{4} - 2 z^{3}) (9 x^{4} y^{8} + 6 x^{2} y^{4} z^{3} + 4 z^{6})

= (3 x^{2} y^{4} - 2 z^{3}) (9 x^{4} y^{8} + 6 x^{2} y^{4} z^{3} + 4 z^{6})

f a c t or cos (x) (9 0^{\circ})

f a c t or (- x) + 2

f a c t or 6 (x + 2 y)^{2} + 13 (x + 2 y) - 5

f (s) = (d s)^{2}

f a c t or e^{x^{2} - y^{2}} cos (2 x y)