faktorisieren 125d^6a^3+216t^3e^9

Lösung

faktorisieren

125 d^{6} a^{3} + 216 t^{3} e^{9}

(5 d^{2} a + 6 e^{3} t) (25 d^{4} a^{2} - 30 e^{3} d^{2} a t + 36 e^{6} t^{2})

Schritte zur Lösung

125 d^{6} a^{3} + 216 t^{3} e^{9}

Schreibe

125 d^{6} a^{3}

um:

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

Schreibe

216 t^{3} e^{9}

um:

(6 t e^{3})^{3}

= (5 d^{2} a)^{3} + (6 t e^{3})^{3}

Wende Formel zur Summe von dritten Potenzen an:

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

(5 d^{2} a)^{3} + (6 t e^{3})^{3} = (5 d^{2} a + 6 t e^{3}) ((5 d^{2} a)^{2} - 5 d^{2} a 6 t e^{3} + (6 t e^{3})^{2})

= (5 d^{2} a + 6 t e^{3}) ((5 d^{2} a)^{2} - 5 d^{2} a 6 t e^{3} + (6 t e^{3})^{2})

Vereinfache

(5 d^{2} a + 6 t e^{3}) ((5 d^{2} a)^{2} - 5 d^{2} a \cdot 6 t e^{3} + (6 t e^{3})^{2}) : (5 d^{2} a + 6 e^{3} t) (25 d^{4} a^{2} - 30 e^{3} d^{2} a t + 36 e^{6} t^{2})

= (5 d^{2} a + 6 e^{3} t) (25 d^{4} a^{2} - 30 e^{3} d^{2} a t + 36 e^{6} t^{2})

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

f a c t or 25 e^{2} - 16

f a c t or 8^{2} + 8 x - 6

f (x) = (sin (3 x))^{2}

s im pl i f y (\frac{d y}{d x})^{4}