erweitern (1/6 x^2+5/6)^3

Lösung

erweitern

(\frac{1}{6} x^{2} + \frac{5}{6})^{3}

\frac{x ^{6} + 125 + 15 x ^{4} + 75 x ^{2}}{216}

Schritte zur Lösung

(\frac{1}{6} x^{2} + \frac{5}{6})^{3}

Wende Formel für perfekte dritte Potenzen an:

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

(\frac{1}{6} x^{2} + \frac{5}{6})^{3} = (\frac{1}{6} x^{2})^{3} + 3 (\frac{1}{6} x^{2})^{2} \frac{5}{6} + 3 \cdot \frac{1}{6} x^{2} (\frac{5}{6})^{2} + (\frac{5}{6})^{3}

= (\frac{1}{6} x^{2})^{3} + 3 (\frac{1}{6} x^{2})^{2} \frac{5}{6} + 3 \cdot \frac{1}{6} x^{2} (\frac{5}{6})^{2} + (\frac{5}{6})^{3}

Vereinfache

(\frac{1}{6} x^{2})^{3} + 3 (\frac{1}{6} x^{2})^{2} \frac{5}{6} + 3 \cdot \frac{1}{6} x^{2} (\frac{5}{6})^{2} + (\frac{5}{6})^{3} : \frac{x ^{6} + 125 + 15 x ^{4} + 75 x ^{2}}{216}

= \frac{x ^{6} + 125 + 15 x ^{4} + 75 x ^{2}}{216}

e x p an d (3 x + 2 x) d x

s im pl i f y (x^{2} + 6) (7 - x)

e x p an d \frac{4 x - 3}{4 ( x - 3 )}

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

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