faktorisieren m^9-27n^{12}

Lösung

faktorisieren

m^{9} - 27 n^{12}

(m^{3} - 3 n^{4}) (m^{6} + 3 m^{3} n^{4} + 9 n^{8})

Schritte zur Lösung

m^{9} - 27 n^{12}

Wende Exponentenregel an

= (m^{3})^{3} - (3 n^{4})^{3}

Wende Formel zur Differenz von dritten Potenzen an:

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

(m^{3})^{3} - (3 n^{4})^{3} = (m^{3} - 3 n^{4}) ((m^{3})^{2} + m^{3} \cdot 3 n^{4} + (3 n^{4})^{2})

= (m^{3} - 3 n^{4}) ((m^{3})^{2} + m^{3} \cdot 3 n^{4} + (3 n^{4})^{2})

Vereinfache

(m^{3})^{2} + m^{3} \cdot 3 n^{4} + (3 n^{4})^{2} : m^{6} + 3 m^{3} n^{4} + 9 n^{8}

= (m^{3} - 3 n^{4}) (m^{6} + 3 m^{3} n^{4} + 9 n^{8})

x \cdot (100 - (\frac{1}{( 0.01 ) \cdot ( 1.01 ) ^{240}})) = 240000

2 x + 9 + 3 x + x = 6 x + 9

(3 \cdot 5)^{6}

\frac{x}{4} - 8 \leq - 42

f a c t or 2 x^{2} + 2 x - 4