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Beliebt Algebra >

faktorisieren 125a^9b^3-64c^3d^6

Lösung

faktorisieren

125 a^{9} b^{3} - 64 c^{3} d^{6}

Lösung

(5 a^{3} b - 4 c d^{2}) (25 a^{6} b^{2} + 20 a^{3} b c d^{2} + 16 c^{2} d^{4})

Schritte zur Lösung

125 a^{9} b^{3} - 64 c^{3} d^{6}

Schreibe

125 a^{9} b^{3}

um:

(5 a^{3} b)^{3}

Schreibe

64 c^{3} d^{6}

um:

(2^{2} c d^{2})^{3}

= (5 a^{3} b)^{3} - (2^{2} c d^{2})^{3}

Wende Formel zur Differenz von dritten Potenzen an:

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

(5 a^{3} b)^{3} - (2^{2} c d^{2})^{3} = (5 a^{3} b - 2^{2} c d^{2}) ((5 a^{3} b)^{2} + 2^{2} c d^{2} 5 a^{3} b + (2^{2} c d^{2})^{2})

= (5 a^{3} b - 2^{2} c d^{2}) ((2^{2} c d^{2})^{2} + 2^{2} c d^{2} 5 a^{3} b + (5 a^{3} b)^{2})

Vereinfache

(5 a^{3} b - 2^{2} c d^{2}) ((5 a^{3} b)^{2} + 2^{2} c d^{2} \cdot 5 a^{3} b + (2^{2} c d^{2})^{2}) : (5 a^{3} b - 4 c d^{2}) (25 a^{6} b^{2} + 20 a^{3} b c d^{2} + 16 c^{2} d^{4})

= (5 a^{3} b - 4 c d^{2}) (25 a^{6} b^{2} + 20 a^{3} b c d^{2} + 16 c^{2} d^{4})

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