faktorisieren x^9-125y^3z^6

Lösung

faktorisieren

x^{9} - 125 y^{3} z^{6}

(x^{3} - 5 y z^{2}) (x^{6} + 5 x^{3} y z^{2} + 25 y^{2} z^{4})

Schritte zur Lösung

x^{9} - 125 y^{3} z^{6}

Schreibe

125 y^{3} z^{6}

um:

(5 y z^{2})^{3}

= x^{9} - (5 y z^{2})^{3}

Wende Exponentenregel an:

a^{b c} = (a^{b})^{c}

x^{9} = (x^{3})^{3}

= (x^{3})^{3} - (5 y z^{2})^{3}

Wende Formel zur Differenz von dritten Potenzen an:

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

(x^{3})^{3} - (5 y z^{2})^{3} = (x^{3} - 5 y z^{2}) (x^{6} + x^{3} 5 y z^{2} + (5 y z^{2})^{2})

= (x^{3} - 5 y z^{2}) (x^{6} + x^{3} 5 y z^{2} + (5 y z^{2})^{2})

Vereinfache

(x^{3} - 5 y z^{2}) (x^{6} + x^{3} \cdot 5 y z^{2} + (5 y z^{2})^{2}) : (x^{3} - 5 y z^{2}) (x^{6} + 5 x^{3} y z^{2} + 25 y^{2} z^{4})

= (x^{3} - 5 y z^{2}) (x^{6} + 5 x^{3} y z^{2} + 25 y^{2} z^{4})

3 (3 w - 5) = 12

\frac{k}{- 12} \leq - 7

5 m + 12 = m + 3

f a c t or 3 y^{2 + 24 y + 48}

s im pl i f y \frac{( a + b ) ^{3}}{18} \cdot \frac{2}{( a + b ) ^{2}}