Solving $\sqrt[3]{x-3}+\sqrt[3]{1-x}=1$ The Equation
How can I analytically show that there are no real solutions for $\sqrt[3]{x-3}+\sqrt[3]{1-x}=1$?
My attempt
With $u = -x+2$
$\sqrt[3]{u-1}-\sqrt[3]{u+1}=1$
Raising to the power of $3$
$$(u+1)^{2/3}(u-1)^{1/3} - (u+1)^{1/3}(u-1)^{2/3}=1\\(u+1)^{1/3}(u^2-1)^{1/3} - (u-1)^{1/3}(u^2-1)^{1/3}=1\\(u^2-1)^{1/3}\cdot\boxed{\left[(u+1)^{1/3}-(u-1)^{1/3}\right]}=1$$
Raising to the power of $3$:
$$(u^2-1)\cdot\left[3(u+1)^{1/3}(u-1)^{2/3}-3(u+1)^{2/3}(u-1)^{1/3}+2\right]=1\\(u^2-1)\cdot\left[3(u^2-1)^{1/3}(u-1)^{1/3}-3(u+1)^{1/3}(u^2-1)^{1/3}+2\right]=1$$
Thus: $(u^2-1)\cdot\left[3(u^2-1)^{1/3}\boxed{\left[(u-1)^{1/3}-(u+1)^{1/3}\right]}+2\right]=1$
And with $y = (u-1)^{1/3}-(u+1)^{1/3}$, we can say that:
$y^3=-3y(u^2-1)^{1/3}+2$
I am stuck... Any tips for this radical equation?

We have that by $A^3-B^3=(A-B)(A^2+AB+B^2)$
$$\sqrt[3]{x-3}+\sqrt[3]{1-x}=\sqrt[3]{x-3}-\sqrt[3]{x-1}=$$
$$=\frac{-2}{\sqrt[3]{(x-3)^2}+\sqrt[3]{(x-3)(x-1)}+\sqrt[3]{(x-1)^2}}<0$$
indeed the latter is true for $x= 1,2,3$ and for $x\neq 1,2,3$ by AM-GM
$$\sqrt[3]{(x-3)^2}+\sqrt[3]{(x-1)^2} > 2\sqrt[3]{|x-3||x-1|}>0$$
and therefore
$$\sqrt[3]{(x-3)^2}+\sqrt[3]{(x-3)(x-1)}+\sqrt[3]{(x-1)^2} >\sqrt[3]{|x-3||x-1|}>0$$
