faktorisieren (x^2+1)^3+(y^2-1)^3

Lösung

faktorisieren

(x^{2} + 1)^{3} + (y^{2} - 1)^{3}

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

Schritte zur Lösung

(x^{2} + 1)^{3} + (y^{2} - 1)^{3}

Wende Formel zur Summe von dritten Potenzen an:

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

(x^{2} + 1)^{3} + (y^{2} - 1)^{3} = ((x^{2} + 1) + (y^{2} - 1)) ((x^{2} + 1)^{2} - (x^{2} + 1) (y^{2} - 1) + (y^{2} - 1)^{2})

= ((x^{2} + 1) + (y^{2} - 1)) ((x^{2} + 1)^{2} - (x^{2} + 1) (y^{2} - 1) + (y^{2} - 1)^{2})

Fasse zusammen

= (x^{2} + y^{2}) ((x^{2} + 1)^{2} - (x^{2} + 1) (y^{2} - 1) + (y^{2} - 1)^{2})

(x^{2} + 1) (y^{2} - 1) = (y + 1) (y - 1) (x^{2} + 1)

(y^{2} - 1)^{2} = (y + 1)^{2} (y - 1)^{2}

= (x^{2} + y^{2}) ((x^{2} + 1)^{2} - (y + 1) (y - 1) (x^{2} + 1) + (y + 1)^{2} (y - 1)^{2})

Multipliziere aus

(x^{2} + 1)^{2} - (y + 1) (y - 1) (x^{2} + 1) + (y + 1)^{2} (y - 1)^{2} : x^{4} + 3 x^{2} - x^{2} y^{2} + y^{4} + 3 - 3 y^{2}

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

\frac{7 c + 3}{2} \leq - 5

x + 6 > 3 x

f = - 3^{2} - 3

- w + 4 (w + 3) = - 12

s im pl i f y (\frac{1}{2}) 4