If the two roots of the equation, $$(a-1)(x^4 + x^2 + 1) + (a+1)(x^2 + x + 1)^2 = 0$$ are real and distinct, then the set of all values of $$a$$ is equal to

We are given the equation $$(a-1)(x^4 + x^2 + 1) + (a+1)(x^2 + x + 1)^2 = 0$$ and told that it has two real and distinct roots. We need to find the set of all values of $$a$$.

First, we simplify the equation. Notice that $$x^4 + x^2 + 1$$ can be factored. Rewrite it as:

$$x^4 + x^2 + 1 = x^4 + 2x^2 + 1 - x^2 = (x^4 + 2x^2 + 1) - x^2 = (x^2 + 1)^2 - x^2.$$

This is a difference of squares:

$$(x^2 + 1)^2 - x^2 = (x^2 + 1 - x)(x^2 + 1 + x) = (x^2 - x + 1)(x^2 + x + 1).$$

So, $$x^4 + x^2 + 1 = (x^2 - x + 1)(x^2 + x + 1)$$. Substitute this into the original equation:

$$(a-1) \cdot (x^2 - x + 1)(x^2 + x + 1) + (a+1) \cdot (x^2 + x + 1)^2 = 0.$$

Factor out $$(x^2 + x + 1)$$:

$$(x^2 + x + 1) \left[ (a-1)(x^2 - x + 1) + (a+1)(x^2 + x + 1) \right] = 0.$$

This equation holds if either $$x^2 + x + 1 = 0$$ or $$(a-1)(x^2 - x + 1) + (a+1)(x^2 + x + 1) = 0$$.

Now, solve $$x^2 + x + 1 = 0$$. The discriminant is $$1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 < 0$$, so it has no real roots. Therefore, the real roots must come from the other equation:

$$(a-1)(x^2 - x + 1) + (a+1)(x^2 + x + 1) = 0.$$

Expand the terms:

$$(a-1)(x^2 - x + 1) = (a-1)x^2 - (a-1)x + (a-1),$$

$$(a+1)(x^2 + x + 1) = (a+1)x^2 + (a+1)x + (a+1).$$

Add them together:

$$(a-1)x^2 - (a-1)x + (a-1) + (a+1)x^2 + (a+1)x + (a+1) = 0.$$

Combine like terms:

• Coefficient of $$x^2$$: $$(a-1) + (a+1) = a - 1 + a + 1 = 2a$$,
• Coefficient of $$x$$: $$-(a-1) + (a+1) = -a + 1 + a + 1 = 2$$,
• Constant term: $$(a-1) + (a+1) = a - 1 + a + 1 = 2a$$.

So the equation becomes:

$$2a x^2 + 2x + 2a = 0.$$

Divide the entire equation by 2 (since 2 ≠ 0):

$$a x^2 + x + a = 0.$$

This is a quadratic equation in $$x$$. For it to have two real and distinct roots, the discriminant must be positive and the leading coefficient must not be zero (to ensure it is quadratic).

The discriminant $$D = b^2 - 4ac$$, where $$a = a$$, $$b = 1$$, $$c = a$$:

$$D = (1)^2 - 4 \cdot a \cdot a = 1 - 4a^2.$$

For two real and distinct roots, $$D > 0$$:

$$1 - 4a^2 > 0,$$

$$4a^2 < 1,$$

$$a^2 < \frac{1}{4},$$

$$-\frac{1}{2} < a < \frac{1}{2}.$$

Additionally, for the equation to be quadratic, the leading coefficient $$a$$ must not be zero. If $$a = 0$$, the equation becomes $$x = 0$$, which has only one root, not two distinct roots. So $$a \neq 0$$.

Therefore, $$a$$ must be in $$\left(-\frac{1}{2}, 0\right) \cup \left(0, \frac{1}{2}\right)$$.

We must also ensure that these roots are valid for the original equation. The factor $$(x^2 + x + 1)$$ has no real roots, so the roots from $$a x^2 + x + a = 0$$ are the only real roots. Since we require exactly two real and distinct roots, and the quadratic provides that under the above conditions, this is sufficient.

Now, comparing with the options:

• A. $$\left(0, \frac{1}{2}\right)$$
• B. $$\left(-\frac{1}{2}, 0\right) \cup \left(0, \frac{1}{2}\right)$$
• C. $$(-\infty, -2) \cup (2, \infty)$$
• D. $$\left(-\frac{1}{2}, 0\right)$$

Option B matches our solution.

Hence, the correct answer is Option B.