faktorisieren 8s^3t^6+125q^3

Lösung

faktorisieren

8 s^{3} t^{6} + 125 q^{3}

(2 s t^{2} + 5 q) (4 s^{2} t^{4} - 10 s t^{2} q + 25 q^{2})

Schritte zur Lösung

8 s^{3} t^{6} + 125 q^{3}

Schreibe

8 s^{3} t^{6}

um:

(2 s t^{2})^{3}

Schreibe

125 q^{3}

um:

(5 q)^{3}

= (2 s t^{2})^{3} + (5 q)^{3}

Wende Formel zur Summe von dritten Potenzen an:

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

(2 s t^{2})^{3} + (5 q)^{3} = (2 s t^{2} + 5 q) ((2 s t^{2})^{2} - 2 s t^{2} 5 q + (5 q)^{2})

= (2 s t^{2} + 5 q) ((2 s t^{2})^{2} - 2 s t^{2} 5 q + (5 q)^{2})

Vereinfache

(2 s t^{2} + 5 q) ((2 s t^{2})^{2} - 2 s t^{2} \cdot 5 q + (5 q)^{2}) : (2 s t^{2} + 5 q) (4 s^{2} t^{4} - 10 s t^{2} q + 25 q^{2})

= (2 s t^{2} + 5 q) (4 s^{2} t^{4} - 10 s t^{2} q + 25 q^{2})

f a c t or 10 - 25

f a c t or - 4 + (- 6 p)

f a c t or 216 - x^{15}

f (x) = (cos (8 x))^{2}

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