faktorisieren 1/27 x^3+1/125 y^3

Lösung

faktorisieren

\frac{1}{27} x^{3} + \frac{1}{125} y^{3}

(\frac{x}{3} + \frac{y}{5}) (\frac{x ^{2}}{9} - \frac{x y}{15} + \frac{y ^{2}}{25})

Schritte zur Lösung

\frac{1}{27} x^{3} + \frac{1}{125} y^{3}

Faktorisiere

\frac{x ^{3}}{27} + \frac{y ^{3}}{125} : (\frac{x}{3} + \frac{y}{5}) ((\frac{1}{3})^{2} x^{2} - \frac{1}{5} \cdot \frac{1}{3} x y + (\frac{1}{5})^{2} y^{2})

= (\frac{x}{3} + \frac{y}{5}) ((\frac{1}{3})^{2} x^{2} - \frac{1}{5} \cdot \frac{1}{3} x y + (\frac{1}{5})^{2} y^{2})

Vereinfache

(\frac{x}{3} + \frac{y}{5}) ((\frac{1}{3})^{2} x^{2} - \frac{1}{5} \cdot \frac{1}{3} x y + (\frac{1}{5})^{2} y^{2}) : (\frac{x}{3} + \frac{y}{5}) (\frac{x ^{2}}{9} - \frac{x y}{15} + \frac{y ^{2}}{25})

= (\frac{x}{3} + \frac{y}{5}) (\frac{x ^{2}}{9} - \frac{x y}{15} + \frac{y ^{2}}{25})

f a c t or 3 x^{3} - 24 x^{2} + 45 x

2 x + 6 = x + 14

b^{2} + 9 b - 14 = 6 b

x^{2} - x + 3 = 2 x^{2} - 5 x - 29

f a c t or 12 m n^{2} + 6 m^{2} n + 2 m