faktorisieren 64a^6-729b^6

Lösung

faktorisieren

64 a^{6} - 729 b^{6}

(2 a + 3 b) (4 a^{2} - 6 ab + 9 b^{2}) (2 a - 3 b) (4 a^{2} + 6 ab + 9 b^{2})

Schritte zur Lösung

64 a^{6} - 729 b^{6}

64 a^{6} = (8 a^{3})^{2}

729 b^{6} = (27 b^{3})^{2}

= (8 a^{3})^{2} - (27 b^{3})^{2}

Wende Formel zur Differenz von zwei Quadraten an:

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

(8 a^{3})^{2} - (27 b^{3})^{2} = (8 a^{3} + 27 b^{3}) (8 a^{3} - 27 b^{3})

= (8 a^{3} + 27 b^{3}) (8 a^{3} - 27 b^{3})

Faktorisiere

8 a^{3} + 27 b^{3} : (2 a + 3 b) (4 a^{2} - 6 ab + 9 b^{2})

= (2 a + 3 b) (4 a^{2} - 6 ab + 9 b^{2}) (8 a^{3} - 27 b^{3})

Faktorisiere

8 a^{3} - 27 b^{3} : (2 a - 3 b) (4 a^{2} + 6 ab + 9 b^{2})

= (2 a + 3 b) (4 a^{2} - 6 ab + 9 b^{2}) (2 a - 3 b) (4 a^{2} + 6 ab + 9 b^{2})

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

33 \cdot x^{3} - 55 \cdot y^{3}

s im pl i f y \frac{( 25 x ^{2} - 15 x + 30 )}{5}

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

e x p an d (x^{3} - 1)^{4}