<s>[INST] If $(a-1)(x^4+x^2+1)+(a+1)(x^2+x+1)^2 = 0$ are real and distinct, Then set of all values of $a$ 
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.
$\bf{Options::}$ $(a)\;\; \displaystyle \left(0,\frac{1}{2}\right)\;\;\;\; (b)\;\; \displaystyle \left(-\frac{1}{2},0\right)\cup \left(0,\frac{1}{2}\right)\;\;\;\; (c)\;\; \left(-\infty, -2 \right)\cup (2,\infty)\;\;\;\; (d)\;\; \displaystyle \left(-\frac{1}{2},0\right)$

$\bf{My\; Try::}$ We can write above equation as $$\displaystyle \frac{a-1}{a+1} = -\frac{(x^2+x+1)^2}{x^4+x^2+1} = -\left[\frac{x^2+x+1}{x^2-x+1}\right]$$
Now We calculate the range of $$\displaystyle \frac{1}{3}\leq \left[\frac{x^2+x+1}{x^2-x+1}\right]\leq 3$$
So We Get $$\displaystyle -3 \leq -\left[\frac{x^2+x+1}{x^2-x+1}\right]\leq - \frac{1}{3}$$
So We get $$\displaystyle -3 \leq \frac{a-1}{a+1}\leq -\frac{1}{3}\Rightarrow -3 -1 \leq 1-\frac{2}{a+1}-1 \leq -\frac{1}{3}-1$$
So we get $$\displaystyle -4\leq -\frac{2}{a+1}\leq -\frac{4}{3}\Rightarrow \frac{2}{3}\leq \frac{1}{a+1} \leq 2$$
So We get $$\displaystyle a\in \left(-\frac{1}{2}\;,\frac{1}{2}\right)$$.
By from above no answer is matching, plz explain me where I am Wrong.
Thanks
 [/INST] A slightly different method. Seemed easier.
$$(a-1)(x^4+x^2+1)+(a+1)(x^2+x+1)^2=0$$
$$(x^2+x+1)\left[(a-1)(x^2-x+1)+(a+1)(x^2+x+1)\right]=0$$
$x^2+x+1=0$ has no real roots. So
$$(a-1)(x^2-x+1)+(a+1)(x^2+x+1)=0$$
$$2ax^2+2x+2a=0$$
$$ax^2+x+a=0$$
For this to have real and distinct roots, we have $\mathrm{Discriminant}>0$.
$$1-4a^2>0$$
$$a\in\left(-\frac{1}{2},0\right)\cup\left(0,\frac{1}{2}\right)$$
$0$ gets excluded because the above quadratic wouldn't be a quadratic for $a=0$, it would only be a linear with one root, and the question asks for distinct roots.
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