erweitern (x^{2/3}-\sqrt[3]{x})^3

Lösung

erweitern

(x^{\frac{2}{3}} - 3 x)^{3}

x^{2} - 3 x^{\frac{5}{3}} + 3 x^{\frac{4}{3}} - x

Schritte zur Lösung

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

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

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

3 (x^{\frac{2}{3}})^{2} 3 x = 3 x^{\frac{5}{3}}

3 x^{\frac{2}{3}} (3 x)^{2} = 3 x^{\frac{4}{3}}

(3 x)^{3} = x

= x^{2} - 3 x^{\frac{5}{3}} + 3 x^{\frac{4}{3}} - x

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

e x p an d \frac{e ^{- 2 - 2} ( 3 + 2 2 )}{8}

e x p an d \frac{4}{3} (15 - 6 y - 7 = - 11)

e x p an d \frac{x ^{2} - 4 x + 1}{( x - 2 ) ^{2}}

s im pl i f y (2 - i)^{4}