faktorisieren 64x^3y^9-125z^3w^6

Lösung

faktorisieren

64 x^{3} y^{9} - 125 z^{3} w^{6}

(4 x y^{3} - 5 z w^{2}) (16 x^{2} y^{6} + 20 x y^{3} z w^{2} + 25 z^{2} w^{4})

Schritte zur Lösung

64 x^{3} y^{9} - 125 z^{3} w^{6}

Schreibe

64 x^{3} y^{9}

um:

(2^{2} x y^{3})^{3}

Schreibe

125 z^{3} w^{6}

um:

(5 z w^{2})^{3}

= (2^{2} x y^{3})^{3} - (5 z w^{2})^{3}

Wende Formel zur Differenz von dritten Potenzen an:

x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})

(2^{2} x y^{3})^{3} - (5 z w^{2})^{3} = (2^{2} x y^{3} - 5 z w^{2}) ((2^{2} x y^{3})^{2} + 5 z w^{2} 2^{2} x y^{3} + (5 z w^{2})^{2})

= (2^{2} x y^{3} - 5 z w^{2}) ((5 z w^{2})^{2} + 5 z w^{2} 2^{2} x y^{3} + (2^{2} x y^{3})^{2})

Vereinfache

(2^{2} x y^{3} - 5 z w^{2}) ((2^{2} x y^{3})^{2} + 5 z w^{2} \cdot 2^{2} x y^{3} + (5 z w^{2})^{2}) : (4 x y^{3} - 5 z w^{2}) (16 x^{2} y^{6} + 20 x y^{3} z w^{2} + 25 z^{2} w^{4})

= (4 x y^{3} - 5 z w^{2}) (16 x^{2} y^{6} + 20 x y^{3} z w^{2} + 25 z^{2} w^{4})

f a c t or x^{5} - 2 x^{- 2}

f a c t or 3 a x + 3 a y - 2 b x - 2 b y

f a c t or 8 x^{9} y^{6} + z^{15}

f (x) = x - 1

f (t) = (cos (2 t))^{2}