faktorisieren m^6n^3+p^3q^3

Lösung

faktorisieren

m^{6} n^{3} + p^{3} q^{3}

(m^{2} n + pq) (m^{4} n^{2} - m^{2} n pq + p^{2} q^{2})

Schritte zur Lösung

m^{6} n^{3} + p^{3} q^{3}

Schreibe

m^{6} n^{3}

um:

(m^{2} n)^{3}

Schreibe

p^{3} q^{3}

um:

(pq)^{3}

= (m^{2} n)^{3} + (pq)^{3}

Wende Formel zur Summe von dritten Potenzen an:

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

(m^{2} n)^{3} + (pq)^{3} = (m^{2} n + pq) ((m^{2} n)^{2} - m^{2} n pq + (pq)^{2})

= (m^{2} n + pq) ((m^{2} n)^{2} - m^{2} n pq + (pq)^{2})

Vereinfache

(m^{2} n + pq) ((m^{2} n)^{2} - m^{2} n pq + (pq)^{2}) : (m^{2} n + pq) (m^{4} n^{2} - m^{2} n pq + p^{2} q^{2})

= (m^{2} n + pq) (m^{4} n^{2} - m^{2} n pq + p^{2} q^{2})

f a c t or 3 x y - 6 x y

f a c t or (\frac{π}{3}) - 3

f (x) = x^{(3)} + 2 x^{(2)} - 5

f a c t or 13 t^{6} - 39 t^{2}

f a c t or a^{5} + cos^{5} (x)