can anyone solve this problem x^8 -1?

Question

can anyone help me solve x raised to the 8th subtracting 1 from everything. I know it is supposed to have 8 answers two of which are x= 1 and x=-1

I need the imaginary and real and preferably all the steps to get there the equation is x^(8) -1

Answer 1

I assume you mean X^8-1 = 0.
If that is the case, there are 8 answers.
(x^4-1)(x^4+1) = 0
(x^2-1)(x^2+1)(x^4+1) = 0
(x+1)(x-1)(x^2+1)(x^4+1) = 0
There are only two real answers, 1 and -1.
The other 6 roots are imaginary. It appears that two of them are i and -i.
The other ones are (sqrt(i)),(-sqrt(i)),(sqrt(-i)), and (-sqrt(-i)).

Answer 2

You have x^8=1. If you need 8 distinct answers then you need to look into the complex plane.

Suppose that x=re^{it) where i = sqrt(-1). At the same time you can express 1 as
1 e^(i 2 n pi), where n=0, 1, 2, 3, 4, 5, 6, or 7.

You now have

r^8 e^(i 8 t) = 1 e^(i 2 n pi)

So r^8=1 or r=1, and
8t = 2 n pi
t = n/4 pi where n=0, 1, 2, 3, 4, 5, 6, or 7.

The answers are
x = e^(i n/4 pi) = cos(n/4 pi) + i sin(n/4 pi) where n=0, 1, 2, 3, 4, 5, 6, or 7.
Note that when n=0 you get x=1 and when n=4 you get x=-1.

Answer 3

To solve this problem you need to factor this binomial using the difference of two squares:

x^8 - 1 = (x^4 + 1)(x^4 - 1)

Keep factoring and then solve each individual factor for x.

There are 8 solutions to this problem, it's just that some are imaginary, some are real, and some are Quadruple roots.

Answer 4

I assume you mean x^8 - 1 = 0 ...

The only real answers are x = 1 and x = -1.

While it is true that a polynomial equation of degree n has n roots, these roots are not necessarily distinct and are not necessarily real numbers. This particular equation is a good illustration of this fact.

Answer 5

i^4 = 1
i^8 = 1

(-i)^2 = -1
(-i)^4 = 1
(-i)^8 = 1

So, for certain we have at least four distinct roots, 1, -1, i, and -i.

Answer 6

umm... that is an expresion