faktorisieren b^6c^9+d^3

Lösung

faktorisieren

b^{6} c^{9} + d^{3}

(b^{2} c^{3} + d) (b^{4} c^{6} - b^{2} c^{3} d + d^{2})

Schritte zur Lösung

b^{6} c^{9} + d^{3}

Schreibe

b^{6} c^{9}

um:

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

= (b^{2} c^{3})^{3} + d^{3}

Wende Formel zur Summe von dritten Potenzen an:

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

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

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

Vereinfache

(b^{2} c^{3} + d) ((b^{2} c^{3})^{2} - b^{2} c^{3} d + d^{2}) : (b^{2} c^{3} + d) (b^{4} c^{6} - b^{2} c^{3} d + d^{2})

= (b^{2} c^{3} + d) (b^{4} c^{6} - b^{2} c^{3} d + d^{2})

s im pl i f y [t^{2} sin (wt)]

s im pl i f y 12 rs t

f a c t or 9 p + 5 p e^{2}

f (x) = 6 x + 7

s im pl i f y (1.2) m