<s>[INST] 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?
 [/INST] Use the fact that $a^3+b^3+c^3-3abc$ is divisible by $a+b+c$.
Let
$$a=\sqrt[3]{x-3}$$
$$b=\sqrt[3]{1-x}$$
$$c=-1$$
Then you have $a+b+c=0$ so
$$a^3+b^3+c^3=3abc$$ or
$$x-3+1-x -1=-3\sqrt[3]{(x-3)(1-x)}$$
and
$$1=\sqrt[3]{(x-3)(1-x)}$$
So
$$(x-3)(1-x)=1$$
and you get $x=2$. If you substitute you find it does not satisfy the equation.
Indeed, this show that $x-3=2-x=-1$
Now the cube roots of $-1$ are roots of $$x^3+1=(x+1)(x^2-x+1)$$ and if $\alpha$ and $\beta$ are the two non real roots, then $$\alpha+\beta =1.$$
So by taking these non real roots, the equation can be solved.
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