To avoid such difficult calculation we can use concept of nth root of unity which says that if x^n=1 then x=e^(2πki/n) where k=0,1,2,3......(n-1) If you put k gretaer than that then roots will starts repeating and also we can simplify them in much nicer form by euler identity which says e^(ix)=cosx+isinx
The only integer solution is 1, and the others are going to occur in pairs as complex numbers. Whenever you have x^n + .... and you're solving, since polynomials split completely over C, you will get n solutions. An even number of them will be complex, and the rest of them will be real.
x = 1 is one solution. Polynomials split completely over C, so there are guaranteed to be five solutions, and since 5 is prime, they will all be unique, and four of them will be complex, occurring in two pairs.
To avoid such difficult calculation we can use concept of nth root of unity which says that if x^n=1 then x=e^(2πki/n) where k=0,1,2,3......(n-1) If you put k gretaer than that then roots will starts repeating and also we can simplify them in much nicer form by euler identity which says e^(ix)=cosx+isinx
this is not math. it's something else 😂😂
x = 1 and very likley solutions with i .....
The only integer solution is 1, and the others are going to occur in pairs as complex numbers.
Whenever you have x^n + .... and you're solving, since polynomials split completely over C, you will get n solutions. An even number of them will be complex, and the rest of them will be real.
X^5=1=1^5X=1
x = 1 is one solution. Polynomials split completely over C, so there are guaranteed to be five solutions, and since 5 is prime, they will all be unique, and four of them will be complex, occurring in two pairs.