faktorisieren (x^3)/(64)-y^6

Lösung

faktorisieren

\frac{x ^{3}}{64} - y^{6}

(\frac{x}{4} - y^{2}) (\frac{x y ^{2}}{4} + \frac{x ^{2}}{16} + y^{4})

Schritte zur Lösung

\frac{x ^{3}}{64} - y^{6}

Schreibe

\frac{x ^{3}}{64} - y^{6}

um:

(3 \frac{1}{64} x)^{3} - (y^{2})^{3}

= (3 \frac{1}{64} x)^{3} - (y^{2})^{3}

Wende Formel zur Differenz von dritten Potenzen an:

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

(3 \frac{1}{64} x)^{3} - (y^{2})^{3} = (3 \frac{1}{64} x - y^{2}) (3 \frac{1}{64})^{2} x^{2} + 3 \frac{1}{64} x y^{2} + y^{4}

= (3 \frac{1}{64} x - y^{2}) (3 \frac{1}{64})^{2} x^{2} + 3 \frac{1}{64} x y^{2} + y^{4}

Fasse zusammen

= (\frac{x}{4} - y^{2}) (\frac{x y ^{2}}{4} + \frac{x ^{2}}{16} + y^{4})

s im pl i f y [4.7]

f a c t or 64 e^{4} ln (2 e) - 16 e^{4}

f a c t or (x + Δ x)^{3}

f a c t or x y^{3} ln (x) d x - 2 d x

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