faktorisieren c^{12}+8d^6

Lösung

faktorisieren

c^{12} + 8 d^{6}

(c^{4} + 2 d^{2}) (c^{8} - 2 c^{4} d^{2} + 4 d^{4})

Schritte zur Lösung

c^{12} + 8 d^{6}

Wende Exponentenregel an

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

Wende Formel zur Summe von dritten Potenzen an:

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

(c^{4})^{3} + (2 d^{2})^{3} = (c^{4} + 2 d^{2}) ((c^{4})^{2} - c^{4} \cdot 2 d^{2} + (2 d^{2})^{2})

= (c^{4} + 2 d^{2}) ((c^{4})^{2} - c^{4} \cdot 2 d^{2} + (2 d^{2})^{2})

Vereinfache

(c^{4})^{2} - c^{4} \cdot 2 d^{2} + (2 d^{2})^{2} : c^{8} - 2 c^{4} d^{2} + 4 d^{4}

= (c^{4} + 2 d^{2}) (c^{8} - 2 c^{4} d^{2} + 4 d^{4})

\frac{5}{6} x - 3 x + \frac{4}{3} = \frac{5}{3}

∣ 2 - x ∣ \leq \frac{3}{7}

s im pl i f y ln (0.9)

f a c t or 5 x^{2} + 17 x - 12

x^{2} - 2 \cdot 0.818 x - 1 \cdot x + 0.818 \cdot 538 = 0