Prove $x^{4}-x+1=0$ has no solution 
I would like to prove that the following equation has no solution in $\mathbb{R}$
   $$x^{4}-x+1=0$$
my question : could we use  Intermediate Value Theorem to prove it otherways I'm interested in more ways of prove that has no solution in $\mathbb{R}$.
without :
i know that we can prove it by that way :
$x=x^4+1\geq 1$ and $x^4-x+1=x^2(x^2-1)+x^2-x+1>0$

 A: $$
x^4-x+1=(x^2-\tfrac12)^2+x^2-x+\tfrac34=(x^2-\tfrac12)^2+(x-\tfrac12)^2+\tfrac12
$$
is never smaller than $1/2$ and can thus have no real roots.
A: This function has an absolute minimum, attained at $\dfrac{\sqrt[3]2}2$, and this minimum is positive.
A: If $x = x^{4}+1$, then certainly $x \geq 1$ as $x^{4} \geq 0$ for all real $x$. But then $x^{4}  = x^{3}x \geq x$ and $x^{4}+1 \geq x+1 > x$, a contradiction.
A: Using analytical method  we have:
$$
y=x^4-x+1 \qquad  y'=4x^3-1 \qquad y''=12x^2
$$
si we have a stationary point for $x=\sqrt[3]{1/4}$ that is a minimum since $y'' $ is always positive.  
We see that
$$
y(\sqrt[3]{1/4})=1-\frac{3}{4}\sqrt[3]{1/4} >0
$$
so, since $y$ is a continuous function $\forall x \in \mathbb{R}$, we have  $y(x)> 0 \forall x \in \mathbb{R}$, and the function has no zeros. 
A general way to find if a polynomial function has real roots, without using analysis and that can be applied in any case, is using the Sturm's theorem. Note that this method can be applied in any case, without searching the roots, i.e. without really solving the polynomial equation,  but it can be extremely laborious .
