erweitern (2^{1/2}+1)^3

Lösung

erweitern

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

5 \cdot 2^{\frac{1}{2}} + 7

Dezimale

14.07106 \dots

Schritte zur Lösung

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

Wende Formel für perfekte dritte Potenzen an:

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

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

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

(2^{\frac{1}{2}})^{3} + 3 (2^{\frac{1}{2}})^{2} \cdot 1 + 3 \cdot 2^{\frac{1}{2}} \cdot 1^{2} + 1^{3} = 5 \cdot 2^{\frac{1}{2}} + 7

= 5 \cdot 2^{\frac{1}{2}} + 7

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