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x²=x
x²-x=0
x(x-1)=0
x=0 ∨ x=1
2]
x²+4=0 brak pierwiastków
3]
x³+1=0
(x+1) (x²-x+1)=0
x=-1
4]
125^2/3=∛125²=∛(5³)²=∛(5²)³=5²=25
125-2/3=124 1/3
5]
x⁶-3x⁴+3x²-1
skorzystam ze wzoru (a-b)³=a³-3a²b+3ab²-b³
czyli nasze b=1
3ab²=3a*1²=3a
czyli 3a=3x² czyli a=x²
więc wielomian ma postac;
(x²-1)³=(x²-1)(x²-1)(x²-1)=(x+1)(x-1)(x+1)(x-1)(x+1)(x-1)=(x+1)³(x-1)³