What is the solution and magnitude of the following problem:

(x^4)-1=0

(x^3)+1=0

(x^3)-1=0

(x^6)-1=0

(x^6)+1=0

x^4 = 1 = 1 cis 2π

x = 1 cis 2π/4 = 1cis π/2

Now add multiples of 2π/4 to that till you get all 4 roots.
You can verify, knowing that 1^4 = (-1)^4 = i^4 = (-i)^4 = 1

Do the others in like wise.

To find the solutions to the given equations, we need to solve for the value(s) of x that make each equation true. Let's solve them step-by-step:

1. (x^4) - 1 = 0

Step 1: Add 1 to both sides of the equation:
(x^4) = 1

Step 2: Take the fourth root of both sides of the equation:
x = ±√1

Solution: There are two solutions: x = 1 and x = -1.

2. (x^3) + 1 = 0

Step 1: Subtract 1 from both sides of the equation:
(x^3) = -1

Step 2: Take the cube root of both sides of the equation:
x = -∛1

Solution: There is one solution: x = -1.

3. (x^3) - 1 = 0

Step 1: Add 1 to both sides of the equation:
(x^3) = 1

Step 2: Take the cube root of both sides of the equation:
x = ∛1

Solution: There is one solution: x = 1.

4. (x^6) - 1 = 0

Step 1: Add 1 to both sides of the equation:
(x^6) = 1

Step 2: Take the sixth root of both sides of the equation:
x = ±√1

Solution: There are two solutions: x = 1 and x = -1.

5. (x^6) + 1 = 0

This equation does not have any real solutions because x raised to an even power will always be positive, and adding 1 cannot make it equal to zero.

Solution: There are no real solutions to this equation.

To find the solutions and magnitude of these equations, we can use algebraic methods. Let's solve each equation one by one.

Equation 1: (x^4) - 1 = 0
To solve this equation, we can rearrange it to isolate x:
(x^4) = 1
Taking the fourth root of both sides, we get:
x = ±1

So, the solutions to this equation are x = 1 and x = -1. The magnitude of these solutions is 1.

Equation 2: (x^3) + 1 = 0
To solve this equation, we can rearrange it to isolate x:
(x^3) = -1
Taking the cube root of both sides, we get:
x = -1

The solution to this equation is x = -1. The magnitude of this solution is 1.

Equation 3: (x^3) - 1 = 0
To solve this equation, we can rearrange it to isolate x:
(x^3) = 1
Taking the cube root of both sides, we get:
x = 1

The solution to this equation is x = 1. The magnitude of this solution is 1.

Equation 4: (x^6) - 1 = 0
This equation is a quadratic equation in terms of (x^3). By factoring, we can solve it:
((x^3) - 1) * ((x^3) + 1) = 0
(x^3) - 1 = 0 or (x^3) + 1 = 0

We have already found the solutions to both of these equations previously. Thus, the solutions to this equation are x = 1, x = -1. The magnitude of these solutions is 1.

Equation 5: (x^6) + 1 = 0
This equation is also a quadratic equation in terms of (x^3). By factoring, we can solve it:
((x^3) - i) * ((x^3) + i) = 0

This equation involves complex numbers, and solving it requires concepts from complex analysis. The solutions involve complex numbers with different magnitudes and directions depending on the values of x. The magnitude cannot be determined without knowing the specific values of x.

In summary:
Equation 1: Solutions: x = 1, x = -1 Magnitude: 1
Equation 2: Solutions: x = -1 Magnitude: 1
Equation 3: Solutions: x = 1 Magnitude: 1
Equation 4: Solutions: x = 1, x = -1 Magnitude: 1
Equation 5: Solutions: Complex numbers Magnitude: Not determinable without specific values of x.