If `x^(3)-x=0 and x^(2)+x^(2)=2`, what is x?

Answer

To solve the equations \( x^3 - x = 0 \) and \( x^2 + x^2 = 2 \), we will follow these steps:

### Step 1: Solve the first equation \( x^3 - x = 0 \)

1. Factor out \( x \):
\[ x(x^2 - 1) = 0 \]

2. Set each factor to zero:
- \( x = 0 \)
- \( x^2 - 1 = 0 \)

3. Solve \( x^2 - 1 = 0 \):
\[ x^2 = 1 \]
Taking the square root gives:
\[ x = 1 \quad \text{or} \quad x = -1 \]

So from the first equation, the possible values of \( x \) are:
\[ x = 0, \quad x = 1, \quad x = -1 \]

### Step 2: Solve the second equation \( x^2 + x^2 = 2 \)

1. Simplify the equation:
\[ 2x^2 = 2 \]

2. Divide both sides by 2:
\[ x^2 = 1 \]

3. Taking the square root gives:
\[ x = 1 \quad \text{or} \quad x = -1 \]

### Step 3: Find the intersection of the solutions

From the first equation, we found:
\[ x = 0, \quad x = 1, \quad x = -1 \]
From the second equation, we found:
\[ x = 1, \quad x = -1 \]

The common solutions (intersection) from both equations are:
\[ x = 1 \quad \text{and} \quad x = -1 \]

### Final Answer:
The values of \( x \) that satisfy both equations are:
\[ x = 1 \quad \text{and} \quad x = -1 \]

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