faktorisieren 125n^9p^6-27s^6f^3

Lösung

faktorisieren

125 n^{9} p^{6} - 27 s^{6} f^{3}

(5 n^{3} p^{2} - 3 s^{2} f) (25 n^{6} p^{4} + 15 n^{3} p^{2} s^{2} f + 9 s^{4} f^{2})

Schritte zur Lösung

125 n^{9} p^{6} - 27 s^{6} f^{3}

Schreibe

125 n^{9} p^{6}

um:

(5 n^{3} p^{2})^{3}

Schreibe

27 s^{6} f^{3}

um:

(3 s^{2} f)^{3}

= (5 n^{3} p^{2})^{3} - (3 s^{2} f)^{3}

Wende Formel zur Differenz von dritten Potenzen an:

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

(5 n^{3} p^{2})^{3} - (3 s^{2} f)^{3} = (5 n^{3} p^{2} - 3 s^{2} f) ((5 n^{3} p^{2})^{2} + 5 n^{3} p^{2} 3 s^{2} f + (3 s^{2} f)^{2})

= (5 n^{3} p^{2} - 3 s^{2} f) ((5 n^{3} p^{2})^{2} + 5 n^{3} p^{2} 3 s^{2} f + (3 s^{2} f)^{2})

Vereinfache

(5 n^{3} p^{2} - 3 s^{2} f) ((5 n^{3} p^{2})^{2} + 5 n^{3} p^{2} \cdot 3 s^{2} f + (3 s^{2} f)^{2}) : (5 n^{3} p^{2} - 3 s^{2} f) (25 n^{6} p^{4} + 15 n^{3} p^{2} s^{2} f + 9 s^{4} f^{2})

= (5 n^{3} p^{2} - 3 s^{2} f) (25 n^{6} p^{4} + 15 n^{3} p^{2} s^{2} f + 9 s^{4} f^{2})

f a c t or (\frac{3 π}{2}) + 2 π

f (x) = sin^{(2)} (x) + 6 x - 2

f (X) = [X + 4]

f a c t or 3 x^{2} + (2 x - 5)

f a c t or \frac{e ^{3} - 1}{3 e}