erweitern (3a+b^2)^4

Lösung

erweitern

(3 a + b^{2})^{4}

81 a^{4} + 108 a^{3} b^{2} + 54 a^{2} b^{4} + 12 a b^{6} + b^{8}

Schritte zur Lösung

(3 a + b^{2})^{4}

Wende Binomialsetz an:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = 3 a, b = b^{2}

= i = 0 \sum 4 (i 4) (3 a)^{(4 - i)} (b^{2})^{i}

Erweitere Summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} (3 a)^{4} (b^{2})^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (3 a)^{3} (b^{2})^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (3 a)^{2} (b^{2})^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (3 a)^{1} (b^{2})^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (3 a)^{0} (b^{2})^{4}

Vereinfache

\frac{4 !}{0 ! ( 4 - 0 ) !} (3 a)^{4} (b^{2})^{0} : 81 a^{4}

Vereinfache

\frac{4 !}{1 ! ( 4 - 1 ) !} (3 a)^{3} (b^{2})^{1} : 108 a^{3} b^{2}

Vereinfache

\frac{4 !}{2 ! ( 4 - 2 ) !} (3 a)^{2} (b^{2})^{2} : 54 a^{2} b^{4}

Vereinfache

\frac{4 !}{3 ! ( 4 - 3 ) !} (3 a)^{1} (b^{2})^{3} : 12 a b^{6}

Vereinfache

\frac{4 !}{4 ! ( 4 - 4 ) !} (3 a)^{0} (b^{2})^{4} : b^{8}

= 81 a^{4} + 108 a^{3} b^{2} + 54 a^{2} b^{4} + 12 a b^{6} + b^{8}

s im pl i f y \frac{( 3 x ) ^{\frac{1}{2}} \cdot ( 12 x ) ^{\frac{3}{2}}}{4}

s im pl i f y (a + 2 b)^{3} + (a - 2 b)^{3}

e x p an d (1 - 2 a)^{3}

s im pl i f y 55 x^{2} - 3 x + 2582 x^{2} - 9 x^{2615} - 3 x + v + 6

e x p an d (3 x - 7)^{3}