faktorisieren 8x^9-125y^3z^6

Lösung

faktorisieren

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

(2 x^{3} - 5 y z^{2}) (4 x^{6} + 10 x^{3} y z^{2} + 25 y^{2} z^{4})

Schritte zur Lösung

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

Schreibe

8 x^{9}

um:

(2 x^{3})^{3}

Schreibe

125 y^{3} z^{6}

um:

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

= (2 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})

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

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

Vereinfache

(2 x^{3} - 5 y z^{2}) ((2 x^{3})^{2} + 2 x^{3} \cdot 5 y z^{2} + (5 y z^{2})^{2}) : (2 x^{3} - 5 y z^{2}) (4 x^{6} + 10 x^{3} y z^{2} + 25 y^{2} z^{4})

= (2 x^{3} - 5 y z^{2}) (4 x^{6} + 10 x^{3} y z^{2} + 25 y^{2} z^{4})

7 - 3 (6 c + 14) > 5 (1 - 2 c)

9 x^{2} + 31 x + 12 \leq 0

\frac{11 - 5 x}{8} - \frac{6 x - 3}{4} = \frac{4 x - 1}{8}

15 t^{2} + 5 t = 0

\frac{( x + 1 ) ( x - 1 )}{x} \leq 0