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Beliebt Algebra >

faktorisieren 27m^9n^6-8a^3b^3

Lösung

faktorisieren

27 m^{9} n^{6} - 8 a^{3} b^{3}

Lösung

(3 m^{3} n^{2} - 2 ab) (9 m^{6} n^{4} + 6 a m^{3} n^{2} b + 4 a^{2} b^{2})

Schritte zur Lösung

27 m^{9} n^{6} - 8 a^{3} b^{3}

Schreibe

27 m^{9} n^{6}

um:

(3 m^{3} n^{2})^{3}

Schreibe

8 a^{3} b^{3}

um:

(2 ab)^{3}

= (3 m^{3} n^{2})^{3} - (2 ab)^{3}

Wende Formel zur Differenz von dritten Potenzen an:

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

(3 m^{3} n^{2})^{3} - (2 ab)^{3} = (3 m^{3} n^{2} - 2 ab) ((3 m^{3} n^{2})^{2} + 3 m^{3} n^{2} 2 ab + (2 ab)^{2})

= (3 m^{3} n^{2} - 2 ab) ((3 m^{3} n^{2})^{2} + 3 m^{3} n^{2} 2 ab + (2 ab)^{2})

Vereinfache

(3 m^{3} n^{2} - 2 ab) ((3 m^{3} n^{2})^{2} + 3 m^{3} n^{2} \cdot 2 ab + (2 ab)^{2}) : (3 m^{3} n^{2} - 2 ab) (9 m^{6} n^{4} + 6 a m^{3} n^{2} b + 4 a^{2} b^{2})

= (3 m^{3} n^{2} - 2 ab) (9 m^{6} n^{4} + 6 a m^{3} n^{2} b + 4 a^{2} b^{2})

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