faktorisieren sin^6(a)+cos^6(a)

Lösung

faktorisieren

sin^{6} (a) + cos^{6} (a)

(sin^{2} (a) + cos^{2} (a)) (sin^{4} (a) - cos^{2} (a) sin^{2} (a) + cos^{4} (a))

Schritte zur Lösung

sin^{6} (a) + cos^{6} (a)

Schreibe

sin^{6} (a) + cos^{6} (a)

um:

(sin^{2} (a))^{3} + (cos^{2} (a))^{3}

= (sin^{2} (a))^{3} + (cos^{2} (a))^{3}

Wende Formel zur Summe von dritten Potenzen an:

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

(sin^{2} (a))^{3} + (cos^{2} (a))^{3} = (sin^{2} (a) + cos^{2} (a)) (sin^{4} (a) - cos^{2} (a) sin^{2} (a) + cos^{4} (a))

= (cos^{2} (a) + sin^{2} (a)) (cos^{4} (a) - cos^{2} (a) sin^{2} (a) + sin^{4} (a))

f a c t or (x + y)^{2} + 4 (x + y) + 4

f a c t or (x + Δ x)^{3} - x^{3}

f a c t or 2 (x cos (ln (x)))

f a c t or 2^{2 x} - 2^{x + 1} + 1

f a c t or 4 wt + 2 w h + 6 i t + 3 ih