faktorisieren 8x^6+27y^3z^9

Lösung

faktorisieren

8 x^{6} + 27 y^{3} z^{9}

(2 x^{2} + 3 y z^{3}) (4 x^{4} - 6 x^{2} y z^{3} + 9 y^{2} z^{6})

Schritte zur Lösung

8 x^{6} + 27 y^{3} z^{9}

Schreibe

8 x^{6}

um:

(2 x^{2})^{3}

Schreibe

27 y^{3} z^{9}

um:

(3 y z^{3})^{3}

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

Wende Formel zur Summe von dritten Potenzen an:

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

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

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

Vereinfache

(2 x^{2} + 3 y z^{3}) ((2 x^{2})^{2} - 3 y z^{3} \cdot 2 x^{2} + (3 y z^{3})^{2}) : (2 x^{2} + 3 y z^{3}) (4 x^{4} - 6 x^{2} y z^{3} + 9 y^{2} z^{6})

= (2 x^{2} + 3 y z^{3}) (4 x^{4} - 6 x^{2} y z^{3} + 9 y^{2} z^{6})

f a c t or r^{2} e^{r x} + r e^{r x} - 2 e^{r x}

f a c t or (5 \cdot (7502 5^{2}) - 4)

f a c t or (4 i y - 54 x)^{7}

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

f a c t or 6 - 32