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.
 A: 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 $\,p_1(x)>4/3.\,$
A: We may observe that our polynomial has the form
$$ (x^2+x+1)\left(x^3 - \frac{5}{2}\right)^2 + \frac{1}{4} (3x^2+3x-1) $$
The first term is always non-negative (and positive for $x \neq \sqrt[3]{\frac{5}{2}}$, that case we can check separately), and the second one is non-negative outside interval $(-2,1)$, so we just need to check the values inside this interval. But $x^2+x+1 = (x+ \frac{1}{2})^2) + \frac{3}{4} \geqslant \frac{3}{4}$, and $(x^3-\frac{5}{2})^2$ has on this interval values not less than $\frac{9}{4}$. It suffices to show that minimum of $\frac{1}{4}(3x^2+3x-1)$ is greater than $\frac{27}{4}$, which is easy to check with elementary methods.
A: I came up with a weird proof, but the idea behind is kind of simple: the fact is that I expect that for $x$ outside an interval the terms $x^8+x^7+x^6$ will be greater than the negative terms, and in this range the remaining terms will be positive. On the other side, for small $x$ the -5 will paly his role and make this negative.
I don't know if this actually happens, but this led me to factorize in this way:
$$ (x^3-5)(x^5+x^4+x^3) + 7x^2+7x+6 = 0$$
Now, continuing with algebraic manipulation:
$$ (x^3-5)x^3(x^2+x+1) + 7(x^2+x+1) = 1 $$
$$ (x^6 - 5x^3 + 7)(x^2+x+1) = 1 $$
The nice thing here is that we have two big terms and they have to multiplicate to 1. Weird. Let us calculate the minima of the two terms. Imposing derivatives equal to zero, we see that the minima of the first is reached in $ x^3 =(15/6)$, which yields
$$ \frac{225}{36} - \frac{450}{36} + 7 = - \frac{225}{36} + \frac{252}{36} = \frac{27}{36} = \frac{3}{4} $$
The second reaches its minimus at $x= -1/2$, which yields $ 1/4 -1/2+1 = 3/4$.
The strategy now is the following: it is enough to show that when one term is $\le 1$, the other is $> 4/3$: in this way the product will be always $>1$.
Case 1. $x^6-5x^3+7 \le 1$. Substituting $t=x^3$, this is equivalent to
$$ (t-3)(t-2) \le 0$$
that is $2 \le t \le 3$. Substituing back $t=x^3$ we get $\sqrt[3]{2} \le x \le \sqrt[3]{3}$. Since the derivative of $x^2+x+1$ is positive for $x \ge 1$, in the above interval it will yield at least
$$ \sqrt[3]{4} +\sqrt[3]{2} +1 \ge 3 > 4/3 $$
Case 2. $x^2+x+1 \le 1$. This is equivalent to $x(x+1) \le 0$, that is $-1 \le x \le 0$. Using again $t=x^3$, we see that $(t-3)(t-2)+1$ is decreasing in the above interval. Thus the minima is attained at $t=0$, for which it is $7 > 4/3$.
