faktorisieren 64c^6+27x^9

Lösung

faktorisieren

64 c^{6} + 27 x^{9}

(4 c^{2} + 3 x^{3}) (9 x^{6} + 16 c^{4} - 12 x^{3} c^{2})

Schritte zur Lösung

64 c^{6} + 27 x^{9}

Wende Exponentenregel an

= (4 c^{2})^{3} + (3 x^{3})^{3}

Wende Formel zur Summe von dritten Potenzen an:

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

(4 c^{2})^{3} + (3 x^{3})^{3} = (4 c^{2} + 3 x^{3}) ((4 c^{2})^{2} - 4 c^{2} \cdot 3 x^{3} + (3 x^{3})^{2})

= (4 c^{2} + 3 x^{3}) ((4 c^{2})^{2} - 4 c^{2} \cdot 3 x^{3} + (3 x^{3})^{2})

Vereinfache

(4 c^{2})^{2} - 4 c^{2} \cdot 3 x^{3} + (3 x^{3})^{2} : 16 c^{4} - 12 x^{3} c^{2} + 9 x^{6}

= (4 c^{2} + 3 x^{3}) (16 c^{4} - 12 x^{3} c^{2} + 9 x^{6})

= (4 c^{2} + 3 x^{3}) (9 x^{6} + 16 c^{4} - 12 x^{3} c^{2})

e x p an d f (x - 4)^{2} + 5

s im pl i f y lo g_{10} (10 a) + lo g_{10} (10 b)

s im pl i f y 0.4 x 1 0^{- 8} 1 x 1 0^{- 1}

f a c t or x^{3} + 4 x^{2} + x \cdot y^{2} + 4 y^{2}

s im pl i f y (4 x y^{2})^{- 3}