erweitern (m^4+1/5 n)^3

Lösung

erweitern

(m^{4} + \frac{1}{5} n)^{3}

\frac{125 m ^{12} + 75 m ^{8} n + n ^{3} + 15 m ^{4} n ^{2}}{125}

Schritte zur Lösung

(m^{4} + \frac{1}{5} n)^{3}

Wende Formel für perfekte dritte Potenzen an:

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

(m^{4} + \frac{1}{5} n)^{3} = (m^{4})^{3} + 3 (m^{4})^{2} \frac{1}{5} n + 3 m^{4} (\frac{1}{5} n)^{2} + (\frac{1}{5} n)^{3}

= (m^{4})^{3} + 3 (m^{4})^{2} \frac{1}{5} n + 3 m^{4} (\frac{1}{5} n)^{2} + (\frac{1}{5} n)^{3}

Vereinfache

(m^{4})^{3} + 3 (m^{4})^{2} \frac{1}{5} n + 3 m^{4} (\frac{1}{5} n)^{2} + (\frac{1}{5} n)^{3} : \frac{125 m ^{12} + 75 m ^{8} n + n ^{3} + 15 m ^{4} n ^{2}}{125}

= \frac{125 m ^{12} + 75 m ^{8} n + n ^{3} + 15 m ^{4} n ^{2}}{125}

e x p an d (2 x + 3 y = 6,) (4 x + 6 y = 12)

e x p an d (2 x y + z^{2}) (3 cos (t))

e x p an d (\frac{7}{x ^{3}} + 5 x^{5})^{5}

s im pl i f y (x) \cdot (x - 4)

s im pl i f y 100.53 (t - sin (t))