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

Lösung

erweitern

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

\frac{x ^{15} - 3 x ^{10} + 3 x ^{5} - 1}{x ^{6}}

Schritte zur Lösung

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

Wende Formel für perfekte dritte Potenzen an:

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

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

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

Vereinfache

(x^{3})^{3} - 3 (x^{3})^{2} \frac{1}{x ^{2}} + 3 x^{3} (\frac{1}{x ^{2}})^{2} - (\frac{1}{x ^{2}})^{3} : \frac{x ^{15} - 3 x ^{10} + 3 x ^{5} - 1}{x ^{6}}

= \frac{x ^{15} - 3 x ^{10} + 3 x ^{5} - 1}{x ^{6}}

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

e x p an d, (2 t^{2} - 1) 2 d t

s im pl i f y (- 3 - 5 i)^{6}

e x p an d \frac{x ^{5} + 2 x ^{3} - 2 x ^{2} - 2 x - 24}{( x - 1 ) ^{2} ( x ^{2} + 4 ) ^{2}}

s im pl i f y (3 a + 5) (3 a + 8)