faktorisieren 125a^6b^3-64c^3

Lösung

faktorisieren

125 a^{6} b^{3} - 64 c^{3}

(5 a^{2} b - 4 c) (25 a^{4} b^{2} + 20 a^{2} b c + 16 c^{2})

Schritte zur Lösung

125 a^{6} b^{3} - 64 c^{3}

Schreibe

125 a^{6} b^{3}

um:

(5 a^{2} b)^{3}

Schreibe

64 c^{3}

um:

(2^{2} c)^{3}

= (5 a^{2} b)^{3} - (2^{2} c)^{3}

Wende Formel zur Differenz von dritten Potenzen an:

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

(5 a^{2} b)^{3} - (2^{2} c)^{3} = (5 a^{2} b - 2^{2} c) ((5 a^{2} b)^{2} + 2^{2} c 5 a^{2} b + (2^{2} c)^{2})

= (5 a^{2} b - 2^{2} c) ((2^{2} c)^{2} + 2^{2} c 5 a^{2} b + (5 a^{2} b)^{2})

Vereinfache

(5 a^{2} b - 2^{2} c) ((5 a^{2} b)^{2} + 2^{2} c \cdot 5 a^{2} b + (2^{2} c)^{2}) : (5 a^{2} b - 4 c) (25 a^{4} b^{2} + 20 a^{2} b c + 16 c^{2})

= (5 a^{2} b - 4 c) (25 a^{4} b^{2} + 20 a^{2} b c + 16 c^{2})

f a c t or H Cl O^{3} + S r (O H) 2

s im pl i f y [0.5]

f a c t or a^{3} + (a + 1)^{3}

\frac{d}{d x} ((f (x)) (x^{3}))

f a c t or x z + yz + x + y