faktorisieren 8a^3b^3-27c^3

Lösung

faktorisieren

8 a^{3} b^{3} - 27 c^{3}

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

Schritte zur Lösung

8 a^{3} b^{3} - 27 c^{3}

Schreibe

8 a^{3} b^{3}

um:

(2 ab)^{3}

Schreibe

27 c^{3}

um:

(3 c)^{3}

= (2 ab)^{3} - (3 c)^{3}

Wende Formel zur Differenz von dritten Potenzen an:

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

(2 ab)^{3} - (3 c)^{3} = (2 ab - 3 c) ((2 ab)^{2} + 3 c 2 ab + (3 c)^{2})

= (2 ab - 3 c) ((3 c)^{2} + 3 c 2 ab + (2 ab)^{2})

Vereinfache

(2 ab - 3 c) ((2 ab)^{2} + 3 c \cdot 2 ab + (3 c)^{2}) : (2 ab - 3 c) (4 a^{2} b^{2} + 6 ab c + 9 c^{2})

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

f a c t or 27 x^{3} y^{3} - 8 z^{6}

f a c t or 2 π r^{2} + 2 π r

f a c t or sin (2 t - 12)

f a c t or - 2 x^{5} + 6

f a c t or (y^{3} - x^{3}) + (y + x)