My solution:
For $x\lt 0$, we have a positive left hand side value and a negative right hand side value. So $x$ can never be a negative value.
For $x\gt 1$, we have:
$1+x^2\gt 2x,\,(1+x^3)(1+x^5)=1+x^3+x^5+x^8\gt 4x^4$ so $(1+x^2)(1+x^3)(1+x^5)\gt 8x^5$, which really is $8x^5\gt 8x^5$, which leads to a contradiction.
For $0\le x \le 1$:
$f(x)=(1+x^2)(1+x^3)(1+x^5)$ has its first derivative of $f'(x)\gt 0$ and so $f$ is an increasing function and so does $f(x)=8x^5$.
That means they can intersect at most once, and by inspection, it is not hard to see that $x=1$ is the only real solution to the system.
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