Prove that $x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6=0$ has no solutions over $\mathbb{R}$

Define the polynomial $$ p(x)\!:=\!x^8\! + \!x^7\! + \!x^6\! - \!5x^5\!- \!5x^4\!- \!5x^3\! + \!7x^2\! + \!7x\!+\!7. \tag{1} $$ By observation it factors as $$ p(x) = (x^2+x+1)(x^6-5x^3+7) = p_1(x)p_2(x^3). \tag{2} $$ If we can prove that for all $\,x\,$ real, $\,p(x)>1\,$ then $\,p(x)-1>0\,$ and hence 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 $\,p_1(x)>4/3.\,$

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.