faktorisieren a^5+b^5c^5

Lösung

faktorisieren

a^{5} + b^{5} c^{5}

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

Schritte zur Lösung

a^{5} + b^{5} c^{5}

Schreibe

b^{5} c^{5}

um:

(b c)^{5}

= a^{5} + (b c)^{5}

Wende Regel zum Faktorisieren an:

x^{n} + y^{n} = (x + y) (x^{n - 1} - x^{n - 2} y + \dots - x y^{n - 2} + y^{n - 1})

n is odd

a^{5} + (b c)^{5} = (a + b c) (a^{4} - a^{3} b c + a^{2} (b c)^{2} - a (b c)^{3} + (b c)^{4})

= (a + b c) (a^{4} - a^{3} b c + a^{2} (b c)^{2} - a (b c)^{3} + (b c)^{4})

Vereinfache

(a + b c) (a^{4} - a^{3} b c + a^{2} (b c)^{2} - a (b c)^{3} + (b c)^{4}) : (a + b c) (a^{4} - a^{3} b c + a^{2} b^{2} c^{2} - a b^{3} c^{3} + b^{4} c^{4})

= (a + b c) (a^{4} - a^{3} b c + a^{2} b^{2} c^{2} - a b^{3} c^{3} + b^{4} c^{4})

2 (x + 8) = 26

s im pl i f y \frac{5 ^{5} e ^{- 5}}{5 !}

2 y^{2} + 11 y + 10 = 0

x^{2} - x + 14 = 0

s im pl i f y \frac{1}{9} - 1