Find the value of `a` if `x^3-3x+a=0` has three distinct real roots.

Explanation

To find the value of \( a \) such that the equation \( x^3 - 3x + a = 0 \) has three distinct real roots, we will follow these steps:

### Step 1: Analyze the function
The function we are dealing with is:
\[ f(x) = x^3 - 3x + a \]
To have three distinct real roots, the function must have two turning points (local maxima and minima) and the function values at these points must be of opposite signs.

### Step 2: Find the critical points
To find the critical points, we differentiate \( f(x) \):
\[ f'(x) = 3x^2 - 3 \]
Setting the derivative equal to zero to find the critical points:
\[ 3x^2 - 3 = 0 \implies x^2 = 1 \implies x = \pm 1 \]
Thus, the critical points are \( x = 1 \) and \( x = -1 \).

### Step 3: Evaluate the function at the critical points
Now, we will evaluate \( f(x) \) at these critical points:
\[ f(1) = 1^3 - 3(1) + a = 1 - 3 + a = a - 2 \]
\[ f(-1) = (-1)^3 - 3(-1) + a = -1 + 3 + a = a + 2 \]

### Step 4: Set conditions for distinct roots
For \( f(x) \) to have three distinct real roots, \( f(1) \) and \( f(-1) \) must have opposite signs:
\[ f(1) \cdot f(-1) < 0 \]
Substituting the values we found:
\[ (a - 2)(a + 2) < 0 \]

### Step 5: Solve the inequality
Now, we will solve the inequality:
\[ a^2 - 4 < 0 \]
This can be factored as:
\[ (a - 2)(a + 2) < 0 \]
The critical points are \( a = -2 \) and \( a = 2 \). The sign of the product changes at these points. We can test intervals:
- For \( a < -2 \): Both factors are negative, product is positive.
- For \( -2 < a < 2 \): One factor is negative, the other is positive, product is negative.
- For \( a > 2 \): Both factors are positive, product is positive.

Thus, the solution to the inequality is:
\[ -2 < a < 2 \]

### Final Answer
The value of \( a \) must satisfy:
\[ a \in (-2, 2) \]