Given information:
- Degree of polynomial: 3
- f(-1) = 10
- f(1) = -6
- f(x) has a critical point at x = -1
- f(x) has a critical point at x = 1

Understanding the problem:
We are given a polynomial function f(x) of degree 3 and some conditions related to its critical points and function values at certain points. We need to determine the x-value of the local minima of the function.

Solution:
To find the local minima of a function, we need to analyze its critical points. Let's break down the solution into steps.

Step 1: Constructing the polynomial function
Since the given function f(x) is a polynomial of degree 3, it can be expressed in the form:

f(x) = ax³ + bx² + cx + d

Step 2: Using the given information
We are given the following information about the function:
- f(-1) = 10
- f(1) = -6
- f(x) has critical points at x = -1 and x = 1

We can use this information to form a system of equations and solve for the coefficients a, b, c, and d. Substituting the x-values into the function, we get:

f(-1) = a(-1)³ + b(-1)² + c(-1) + d = 10
f(1) = a(1)³ + b(1)² + c(1) + d = -6

Simplifying the equations, we have:
-a + b - c + d = 10 ...(1)
a + b + c + d = -6 ...(2)

Step 3: Finding the critical points
To find the critical points, we need to differentiate the function and set the derivative equal to zero. Differentiating f(x), we get:

f'(x) = 3ax² + 2bx + c

Since f(x) has critical points at x = -1 and x = 1, we can set the derivative equal to zero and solve for the coefficients:

f'(-1) = 3a(-1)² + 2b(-1) + c = 0
f'(1) = 3a(1)² + 2b(1) + c = 0

Simplifying the equations, we have:
3a - 2b + c = 0 ...(3)
3a + 2b + c = 0 ...(4)

Step 4: Solving the system of equations
Now we have a system of four equations (1, 2, 3, and 4) with four unknowns (a, b, c, and d). We can solve this system of equations to find the values of a, b, c, and d.

Adding equations (1) and (2), we get:
2b + 2d = 4 ...(5)

Subtracting equation (4) from equation (3), we get:
-4b = 0 ...(6)

From equation (6), we can conclude that b = 0.

Sub