faktorisieren 125x^9-27y^6

Lösung

faktorisieren

125 x^{9} - 27 y^{6}

(5 x^{3} - 3 y^{2}) (25 x^{6} + 15 x^{3} y^{2} + 9 y^{4})

Schritte zur Lösung

125 x^{9} - 27 y^{6}

Wende Exponentenregel an

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

Wende Formel zur Differenz von dritten Potenzen an:

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

(5 x^{3})^{3} - (3 y^{2})^{3} = (5 x^{3} - 3 y^{2}) ((5 x^{3})^{2} + 5 x^{3} \cdot 3 y^{2} + (3 y^{2})^{2})

= (5 x^{3} - 3 y^{2}) ((5 x^{3})^{2} + 5 x^{3} \cdot 3 y^{2} + (3 y^{2})^{2})

Vereinfache

(5 x^{3})^{2} + 5 x^{3} \cdot 3 y^{2} + (3 y^{2})^{2} : 25 x^{6} + 15 x^{3} y^{2} + 9 y^{4}

= (5 x^{3} - 3 y^{2}) (25 x^{6} + 15 x^{3} y^{2} + 9 y^{4})

- 5 < \frac{z}{- 1} - 8

(x - 2) (x + 5) = 0

s im pl i f y (3 x)^{\frac{4}{5}}

- 12 + \frac{n}{10} > - 7

6 (7 x - 1) = 43 x