faktorisieren (b-c)^3+(c-a)^3

Lösung

faktorisieren

(b - c)^{3} + (c - a)^{3}

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

Schritte zur Lösung

(b - c)^{3} + (c - a)^{3}

Wende Formel zur Summe von dritten Potenzen an:

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

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

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

Fasse zusammen

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

Multipliziere aus

(b - c)^{2} - (b - c) (c - a) + (c - a)^{2} : 3 c^{2} - 3 b c - 3 c a + b^{2} + a^{2} + ba

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

2 x^{2} = 13 x + 7

3 x + 6 = 6 x

s im pl i f y - \frac{1.1 ( 5 )}{( 0.01 ( 5 ) ^{2} + 1 ) ^{2}}

3 (x - 1) + 7 > 4 x + 2 - 3 (2 - x)

2 (x - 1) < 1 - 6 x