faktorisieren (m-2)^3+(m+5)^3

Lösung

faktorisieren

(m - 2)^{3} + (m + 5)^{3}

(2 m + 3) (m^{2} + 3 m + 39)

Schritte zur Lösung

(m - 2)^{3} + (m + 5)^{3}

Wende Formel zur Summe von dritten Potenzen an:

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

(m - 2)^{3} + (m + 5)^{3} = ((m - 2) + (m + 5)) ((m - 2)^{2} - (m + 5) (m - 2) + (m + 5)^{2})

= ((m + 5) + (m - 2)) ((m + 5)^{2} - (m + 5) (m - 2) + (m - 2)^{2})

Fasse zusammen

= (2 m + 3) ((m + 5)^{2} - (m + 5) (m - 2) + (m - 2)^{2})

Multipliziere aus

(m - 2)^{2} - (m + 5) (m - 2) + (m + 5)^{2} : m^{2} + 3 m + 39

= (2 m + 3) (m^{2} + 3 m + 39)

f a c t or 2 m x - 20 n y - 8 n x + 5 m y

s im pl i f y [6]

f a c t or - cos (3) (3 x^{2})

f a c t or \frac{e ^{9} - e ^{3}}{6}

f a c t or a x + x^{2} + ab + b x