erweitern (a_{1}e+a_{2}e)^3

Lösung

erweitern

(a_{1} e + a_{2} e)^{3}

e^{3} a_{1}^{3} + 3 e^{3} a_{1}^{2} a_{2} + 3 e^{3} a_{1} a_{2}^{2} + e^{3} a_{2}^{3}

Schritte zur Lösung

(a_{1} e + a_{2} e)^{3}

= (e a_{1} + e a_{2})^{3}

Wende Formel für perfekte dritte Potenzen an:

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

a = a_{1} e, b = a_{2} e

= (a_{1} e)^{3} + 3 (a_{1} e)^{2} a_{2} e + 3 a_{1} e (a_{2} e)^{2} + (a_{2} e)^{3}

Vereinfache

(a_{1} e)^{3} + 3 (a_{1} e)^{2} a_{2} e + 3 a_{1} e (a_{2} e)^{2} + (a_{2} e)^{3} : e^{3} a_{1}^{3} + 3 e^{3} a_{1}^{2} a_{2} + 3 e^{3} a_{1} a_{2}^{2} + e^{3} a_{2}^{3}

= e^{3} a_{1}^{3} + 3 e^{3} a_{1}^{2} a_{2} + 3 e^{3} a_{1} a_{2}^{2} + e^{3} a_{2}^{3}

e x p an d (y^{2} - 9) (y^{2} - 9)^{'}

e x p an d (s + 4) (s + \frac{1}{2}) (s + \frac{6}{5}) (s + \frac{5}{2})

e x p an d (e^{2 x} - 5 e^{x} + 6 = 1) e^{x}

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

e x p an d \frac{s ^{3} ( 5 t ^{2} - 6 t + 3 H ( t ) - 3 e ^{- 2 t} )}{3 ( s + 2 )}