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First, expand both sides:

Left side: $(a-1)(1+x+x^2{)}^2=(a-1)(1+2x+3x^2+2x^3+x^4)$

Right side: $(a+1)(1+x^2+x^4)$

Bring all terms to one side: $$(a-1)(1+2x+3x^2+2x^3+x^4)-(a+1)(1+x^2+x^4)=0$$

Expand: $(a-1)(1)+(a-1)(2x)+(a-1)(3x^2)+(a-1)(2x^3)+(a-1)(x^4)$ $-(a+1)(1)-(a+1)(x^2)-(a+1)(x^4)=0$

Group like terms: - Constant: $(a-1)-(a+1)=-2$ - $x$ term: $2(a-1)$ - $x^2$ term: $3(a-1)-(a+1)=3a-3-a-1=2a-4$ - $x^3$ term: $2(a-1)$ - $x^4$ term: $(a-1)-(a+1)=-2$

So the equation becomes: $$-2+2(a-1)x+(2a-4)x^2+2(a-1)x^3-2x^4=0$$

Divide both sides by $2$: $$-1+(a-1)x+(a-2)x^2+(a-1)x^3-x^4=0$$

Or: $$-x^4+(a-1)x^3+(a-2)x^2+(a-1)x-1=0$$

Or, multiplying both sides by $-1$: $$x^4-(a-1)x^3-(a-2)x^2-(a-1)x+1=0$$

Let us factor this quartic. Notice that if $x=1$:

$x^4-(a-1)x^3-(a-2)x^2-(a-1)x+1=1-(a-1)-(a-2)-(a-1)+1$ $=1-a+1-a+2-a+1+1$ But let's compute step by step: $1-(a-1)=2-a$ $2-a-(a-2)=2-a-a+2=4-2a$ $4-2a-(a-1)=4-2a-a+1=5-3a$ $5-3a+1=6-3a$ So $x=1$ is a root if $6-3a=0$, i.e., $a=2$.

Try $x=-1$: $(-1{)}^4-(a-1)(-1{)}^3-(a-2)(-1{)}^2-(a-1)(-1)+1$ $=1-(a-1)(-1)-(a-2)(1)-(a-1)(-1)+1$ $=1+(a-1)-(a-2)+(a-1)+1$ $=1+a-1-a+2+a-1+1$ $=(1-1+2+1)+(a-a+a)-1$ $=(3)+(a)-1$ $=a+2$ So $x=-1$ is a root if $a=-2$.

Alternatively, factor the quartic as follows:

Let us try to factor as $(x^2+bx+1)(x^2+cx+1)$.

Expand: $(x^2+bx+1)(x^2+cx+1)=x^4+(b+c)x^3+(bc+2)x^2+(b+c)x+1$

Compare with $x^4-(a-1)x^3-(a-2)x^2-(a-1)x+1$:

So: - $x^4$ coefficient: $1$ - $x^3$ coefficient: $b+c=-(a-1)$ - $x^2$ coefficient: $bc+2=-(a-2)$ - $x$ coefficient: $b+c=-(a-1)$ - constant: $1$

So $b+c=-(a-1)$ $bc+2=-(a-2)$

Let $b+c=s=-(a-1)$ $bc=-(a-2)-2=-(a)+2-2=-a$

So $b$ and $c$ are roots of $t^2-st-a=0$

So $t^2+(a-1)t-a=0$

Thus, the quartic factors as: $$(x^2+bx+1)(x^2+cx+1)=0$$ where $b$ and $c$ are roots of $t^2+(a-1)t-a=0$.

Therefore, the solutions are all real $x$ such that $$x^2+bx+1=0\text{or}{x}^2+cx+1=0$$ where $b$ and $c$ are roots of $t^2+(a-1)t-a=0$.

Explicitly, $b,c=\frac{-(a-1)\pm \sqrt{(a-1{)}^2+4a}}{2}=\frac{-(a-1)\pm \sqrt{a^2+2a+1+4a}}{2}=\frac{-(a-1)\pm \sqrt{a^2+6a+1}}{2}$

So the solutions are all real $x$ such that $$x^2+\left(\frac{-(a-1)+\sqrt{a^2+6a+1}}{2}\right)x+1=0$$ or $$x^2+\left(\frac{-(a-1)-\sqrt{a^2+6a+1}}{2}\right)x+1=0$$

That is, for each $a$, the equation has at most four real solutions, given by the roots of these two quadratics.

Special cases: - If $a=2$, then $b+c=-(2-1)=-1$, $bc=-2$; so $b$ and $c$ are roots of $t^2+t-2=0$, i.e., $t=1,-2$. So the quadratics are $x^2+x+1=0$ and $x^2-2x+1=0$. The first has no real roots, the second is $(x-1{)}^2=0$, so $x=1$ is a double root. - If $a=-2$, then $b+c=-(-2-1)=3$, $bc=2$; so $t^2-3t-2=0$, $t=\frac{3\pm \sqrt{9+8}}{2}=\frac{3\pm \sqrt{17}}{2}$. Both quadratics $x^2+bx+1=0$ and $x^2+cx+1=0$ have real roots.

In summary:

For each real $a$, the solutions are all real $x$ such that $$x^2+bx+1=0\text{or}{x}^2+cx+1=0$$ where $b$ and $c$ are the roots of $t^2+(a-1)t-a=0$.