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Beliebt Algebra >

faktorisieren 125x^6y^{12}-64a^6b^{33}

Lösung

faktorisieren

125 x^{6} y^{12} - 64 a^{6} b^{33}

Lösung

(5 x^{2} y^{4} - 4 a^{2} b^{11}) (25 x^{4} y^{8} + 20 a^{2} b^{11} x^{2} y^{4} + 16 a^{4} b^{22})

Schritte zur Lösung

125 x^{6} y^{12} - 64 a^{6} b^{33}

Schreibe

125 x^{6} y^{12}

um:

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

Schreibe

64 a^{6} b^{33}

um:

(2^{2} a^{2} b^{11})^{3}

= (5 x^{2} y^{4})^{3} - (2^{2} a^{2} b^{11})^{3}

Wende Formel zur Differenz von dritten Potenzen an:

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

(5 x^{2} y^{4})^{3} - (2^{2} a^{2} b^{11})^{3} = (5 x^{2} y^{4} - 2^{2} a^{2} b^{11}) ((5 x^{2} y^{4})^{2} + 2^{2} a^{2} b^{11} 5 x^{2} y^{4} + (2^{2} a^{2} b^{11})^{2})

= (5 x^{2} y^{4} - 2^{2} a^{2} b^{11}) ((2^{2} a^{2} b^{11})^{2} + 2^{2} a^{2} b^{11} 5 x^{2} y^{4} + (5 x^{2} y^{4})^{2})

Vereinfache

(5 x^{2} y^{4} - 2^{2} a^{2} b^{11}) ((5 x^{2} y^{4})^{2} + 2^{2} a^{2} b^{11} \cdot 5 x^{2} y^{4} + (2^{2} a^{2} b^{11})^{2}) : (5 x^{2} y^{4} - 4 a^{2} b^{11}) (25 x^{4} y^{8} + 20 a^{2} b^{11} x^{2} y^{4} + 16 a^{4} b^{22})

= (5 x^{2} y^{4} - 4 a^{2} b^{11}) (25 x^{4} y^{8} + 20 a^{2} b^{11} x^{2} y^{4} + 16 a^{4} b^{22})

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