<s>[INST] Find all the real solutions of $(1+x^2)(1+x^4)=4x^3$. As title suggests, the question is to find all the real roots to the polynomial:
$$(1+x^2)(1+x^4)=4x^3$$
This problem was asked in the Kettering University Math Olympiad a few years back, it's an interesting problem with many different ways to approach it. I'm going to share my own approach here, please make sure to let me know if there are any mistakes in mine or if anything can be improved, and share your own methods too!
Here's my approach for the problem:
$$(1+x^2)(1+x^4)=4x^3$$
Divide by $x^3$ on both sides:
$$\frac{1+x^2}{x}\cdot\frac{1+x^4}{x^2}=4$$
$$\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}\right)=4$$
Now we can set $y=x+\frac{1}{x}$, thus, $y^2-2=x^2+\frac{1}{x^2}$
$$y(y^2-2)=4$$
$$y^3-2y-4=0$$
$$y^3-8-2y+4=0$$
$$(y-2)(y^2+2y+4)-2(y-2)=0$$
$$(y-2)(y^2+2y+2)=0$$
Now obviously, since we need to find real solutions, the term $(y^2+2y+2)$ will need to be rejected, as it can easily be checked that it yields no real roots.
[Alternatively, we can set $y^2+2y+2=0$ and observe that $(y+1)^2=-1$, thus this term yields no real solution]
Therefore:
$$y=2$$
$$x+\frac{1}{x}=2$$
$$\frac{x^2+1}{x}=2$$
$$x^2-2x+1=0$$
$$(x-1)^2=0$$
Therefore, $x=1$ is our only real solution.
 [/INST] After expanding, we're trying to find the roots of the polynomial $p(x)=x^6+x^4-4x^3+x^2+1$. Factoring out obvious roots one at time gives
\begin{align*}
p(x)&=x^6+x^4-4x^3+x^2+1\\
&=(x-1)(x^5+x^4+2x^3-2x^2-x-1)\\
&=(x-1)^2(x^4+2x^3+4x^2+2x+1)\\
&=(x-1)^2\left[(x^2+x)^2+2x^2+(x+1)^2\right]
\end{align*}
at which point $x=1$ is clearly the only real root.
</s>