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Beliebt Algebra >

faktorisieren m^3n^6-64p^3

Lösung

faktorisieren

m^{3} n^{6} - 64 p^{3}

Lösung

(m n^{2} - 4 p) (m^{2} n^{4} + 4 m n^{2} p + 16 p^{2})

Schritte zur Lösung

m^{3} n^{6} - 64 p^{3}

Schreibe

m^{3} n^{6}

um:

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

Schreibe

64 p^{3}

um:

(2^{2} p)^{3}

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

Wende Formel zur Differenz von dritten Potenzen an:

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

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

= (m n^{2} - 2^{2} p) ((m n^{2})^{2} + m n^{2} 2^{2} p + (2^{2} p)^{2})

Vereinfache

(m n^{2} - 2^{2} p) ((m n^{2})^{2} + m n^{2} \cdot 2^{2} p + (2^{2} p)^{2}) : (m n^{2} - 4 p) (m^{2} n^{4} + 4 m n^{2} p + 16 p^{2})

= (m n^{2} - 4 p) (m^{2} n^{4} + 4 m n^{2} p + 16 p^{2})

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