erweitern (x^2+1/x)^3

Lösung

erweitern

(x^{2} + \frac{1}{x})^{3}

\frac{x ^{9} + 3 x ^{6} + 3 x ^{3} + 1}{x ^{3}}

Schritte zur Lösung

(x^{2} + \frac{1}{x})^{3}

Wende Formel für perfekte dritte Potenzen an:

(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

(x^{2} + \frac{1}{x})^{3} = (x^{2})^{3} + 3 (x^{2})^{2} \frac{1}{x} + 3 x^{2} (\frac{1}{x})^{2} + (\frac{1}{x})^{3}

= (x^{2})^{3} + 3 (x^{2})^{2} \frac{1}{x} + 3 x^{2} (\frac{1}{x})^{2} + (\frac{1}{x})^{3}

Vereinfache

(x^{2})^{3} + 3 (x^{2})^{2} \frac{1}{x} + 3 x^{2} (\frac{1}{x})^{2} + (\frac{1}{x})^{3} : \frac{x ^{9} + 3 x ^{6} + 3 x ^{3} + 1}{x ^{3}}

= \frac{x ^{9} + 3 x ^{6} + 3 x ^{3} + 1}{x ^{3}}

e x p an d \frac{1}{( 2 + x ) ^{2}}

e x p an d - \frac{3 u ^{2} + 2 u - 2}{u ( u ^{3} - 1 )}

e x p an d 8 cos^{2} (2 t) + (t - 5)^{2} - 7 e^{4 t}

s im pl i f y δ (x - 3)

e x p an d \frac{- x ^{2} - 9}{( x ^{2} - 9 ) ^{2}}