<s>[INST] Prove that $x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6=0$ has no solutions over $\mathbb{R}$ The problem goes as follows:

Using elementary methods prove that $x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6=0$ has no solutions for $x \in \mathbb{R}$.

I first came across the problem in a school book targeted towards students just exposed to factorisation and it intrigued me how one could prove this without Sturm's Theorem. Maybe one could try to factorise into squares, but I can get no result from that. Thanks in advance.
 [/INST] Define the real polynomial
$$ p(x)\!:=\!x^8\! + \!x^7\! + \!x^6\! - \!5x^5\!-
 \!5x^4\!- \!5x^3\! + \!7x^2\! + \!7x\!+\!7. \tag{1} $$
By observation, the polynomial easily factors as the product
$$ p(x) = (x^2+x+1)(x^6-5x^3+7) = p_1(x)p_2(x^3). \tag{2} $$
If we can prove for all real $\,x\,$ that $\,p(x)>1\,$ then
also $\,p(x)-1>0\,$ and hence $\,p(x)-1,\,$ the polynomial
in question, has no real roots.
Note that
$$ p_1(x) := x^2 + x + 1 = (x+1/2)^2+(3/4) \ge 3/4
 \tag{3} $$
and
$$ p_2(x) := x^2-5x+7 = (x-5/2)^2+(3/4) \ge 3/4.
 \tag{4} $$
Note that $\,p_1(x)\le 1\,$ iff $\,-1\le x\le 0\,$ and
$\,p_2(x^3)\le 1\,$ iff $\,2\le x^3\le3\,$ which are disjoint intervals. Now prove that when $\,p_1(x)<1\,$ that $\,p_2(x^3)>4/3\,$ and also when $\,p_2(x^3)<1\,$ that $\,p_1(x)>4/3.\,$
</s>