Let's perform the indicated operations on the given functions $f(x) = x^3 + x^2 + 1$ and $g(x) = -6x^2 + 2$.

1. $(f + g)(x)$

The sum of the functions $f(x)$ and $g(x)$ is:

$(f + g)(x) = f(x) + g(x)$

$(f + g)(x) = (x^3 + x^2 + 1) + (-6x^2 + 2)$

Now, combine like terms:

$(f + g)(x) = x^3 + (x^2 - 6x^2) + (1 + 2)$

$(f + g)(x) = x^3 - 5x^2 + 3$

There are no domain restrictions for this sum since both functions are polynomials, which are defined for all real numbers. So, the domain is $(-\infty, \infty)$.

2. $(f - g)(x)$

The difference of the functions $f(x)$ and $g(x)$ is:

$(f - g)(x) = f(x) - g(x)$

$(f - g)(x) = (x^3 + x^2 + 1) - (-6x^2 + 2)$

Distribute the negative sign:

$(f - g)(x) = x^3 + x^2 + 1 + 6x^2 - 2$

Now, combine like terms:

$(f - g)(x) = x^3 + (x^2 + 6x^2) + (1 - 2)$

$(f - g)(x) = x^3 + 7x^2 - 1$

Again, there are no domain restrictions for this difference since both functions are polynomials. The domain is $(-\infty, \infty)$.

3. $(f \cdot g)(x)$

The product of the functions $f(x)$ and $g(x)$ is:

$(f \cdot g)(x) = f(x) \cdot g(x)$

$(f \cdot g)(x) = (x^3 + x^2 + 1) \cdot (-6x^2 + 2)$

Now, distribute $(-6x^2 + 2)$ across $(x^3 + x^2 + 1)$:

$(f \cdot g)(x) = (x^3 \cdot -6x^2) + (x^3 \cdot 2) + (x^2 \cdot -6x^2) + (x^2 \cdot 2) + (1 \cdot -6x^2) + (1 \cdot 2)$

Simplify each term:

$(f \cdot g)(x) = -6x^5 + 2x^3 - 6x^4 + 2x^2 - 6x^2 + 2$

Combine like terms:

$(f \cdot g)(x) = -6x^5 - 6x^4 + 2x^3 - 4x^2 + 2$

Since both functions are polynomials, the domain is $(-\infty, \infty)$.

4. $\left( \frac{f}{g} \right)(x)$

The division of the functions $f(x)$ and $g(x)$ is:

$\left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} = \frac{x^3 + x^2 + 1}{-6x^2 + 2}$

For this quotient, the domain will be restricted by the values of $x$ that make the denominator zero, because division by zero is undefined.

Set $g(x) = 0$:

$-6x^2 + 2 = 0$

Solve for $x$:

$-6x^2 = -2 \implies x^2 = \frac{1}{3} \implies x = \pm \frac{\sqrt{3}}{3}$

Thus, the function is undefined at $x = \pm \frac{\sqrt{3}}{3}$.

Therefore, the domain of $\left( \frac{f}{g} \right)(x)$ is all real numbers except $x = \pm \frac{\sqrt{3}}{3}$, or:

$\boxed{(-\infty, -\frac{\sqrt{3}}{3}) \cup (-\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}) \cup (\frac{\sqrt{3}}{3}, \infty)}$

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Would you like to explore any part of this solution in more detail?

Here are 5 related questions to extend your learning:

1. How would the behavior of $f(x) + g(x)$ change if one function had a different degree?
2. Can you analyze the behavior of $f(x) \cdot g(x)$ for large values of $x$?
3. What is the importance of domain restrictions when dividing functions?
4. How can you graph $(f + g)(x)$ and $(f - g)(x)$ to visualize their combined effects?
5. How does polynomial degree affect the long-term behavior of products like $f(x) \cdot g(x)$?

Tip: When multiplying or dividing functions, always check for domain restrictions caused by potential zero values in the denominator.