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Beliebt Algebra >

faktorisieren 64w^6z^9-27y^3

Lösung

faktorisieren

64 w^{6} z^{9} - 27 y^{3}

Lösung

(4 w^{2} z^{3} - 3 y) (16 w^{4} z^{6} + 12 w^{2} z^{3} y + 9 y^{2})

Schritte zur Lösung

64 w^{6} z^{9} - 27 y^{3}

Schreibe

64 w^{6} z^{9}

um:

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

Schreibe

27 y^{3}

um:

(3 y)^{3}

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

Wende Formel zur Differenz von dritten Potenzen an:

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

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

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

Vereinfache

(2^{2} w^{2} z^{3} - 3 y) ((2^{2} w^{2} z^{3})^{2} + 2^{2} w^{2} z^{3} \cdot 3 y + (3 y)^{2}) : (4 w^{2} z^{3} - 3 y) (16 w^{4} z^{6} + 12 w^{2} z^{3} y + 9 y^{2})

= (4 w^{2} z^{3} - 3 y) (16 w^{4} z^{6} + 12 w^{2} z^{3} y + 9 y^{2})

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