faktorisieren x^{12}-y^{18}

Lösung

faktorisieren

x^{12} - y^{18}

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

Schritte zur Lösung

x^{12} - y^{18}

x^{12} = (x^{6})^{2}

y^{18} = (y^{9})^{2}

= (x^{6})^{2} - (y^{9})^{2}

Wende Formel zur Differenz von zwei Quadraten an:

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

(x^{6})^{2} - (y^{9})^{2} = (x^{6} + y^{9}) (x^{6} - y^{9})

= (x^{6} + y^{9}) (x^{6} - y^{9})

Faktorisiere

x^{6} + y^{9} : (x^{2} + y^{3}) (x^{4} - x^{2} y^{3} + y^{6})

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

Faktorisiere

x^{6} - y^{9} : (x^{2} - y^{3}) (x^{4} + x^{2} y^{3} + y^{6})

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

s im pl i f y \frac{( 4 x + 4 )}{2}

s im pl i f y \frac{( x ^{\frac{3}{2}} )}{( x ^{- \frac{1}{2}} )}

e x p an d (x + y)^{4} - 1

s im pl i f y (\frac{3}{2} x + \frac{2}{3})^{3}

s im pl i f y v^{2} v^{6} + 4 v^{12}