erweitern (L^{1/3}+K^{1/3})^3

Lösung

erweitern

(L^{\frac{1}{3}} + K^{\frac{1}{3}})^{3}

L + 3 L^{\frac{2}{3}} K^{\frac{1}{3}} + 3 L^{\frac{1}{3}} K^{\frac{2}{3}} + K

Schritte zur Lösung

(L^{\frac{1}{3}} + K^{\frac{1}{3}})^{3}

Wende Formel für perfekte dritte Potenzen an:

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

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

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

(L^{\frac{1}{3}})^{3} = L

(L^{\frac{1}{3}})^{2} = L^{\frac{2}{3}}

(K^{\frac{1}{3}})^{2} = K^{\frac{2}{3}}

(K^{\frac{1}{3}})^{3} = K

= L + 3 L^{\frac{2}{3}} K^{\frac{1}{3}} + 3 L^{\frac{1}{3}} K^{\frac{2}{3}} + K

e x p an d (cos (x) + y sin (x))^{2}

e x p an d P (2, - 3) A (c + 2, - 4) B (4 - 3 c, - 2)

e x p an d l im x \Rightarrow 2 (3 x^{3} - 2 x^{2} + 6 x)

s im pl i f y (5 - x + 2)^{2}

e x p an d 3 (13 + 7)