Solving $\sqrt[3]{x-3}+\sqrt[3]{1-x}=1$

Question

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?

Answer 1

Certainly for any real $x$,

$$x-3 < x-1$$

so

$$\sqrt[3]{x-3} < \sqrt[3]{x-1}$$

since cube root is a monotonically increasing function. So

$$\sqrt[3]{x-3} + \sqrt[3]{1-x} = \sqrt[3]{x-3} - \sqrt[3]{x-1} < 0<1.$$

Answer 2

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.

Answer 3

If you're looking for real solutions, I think you can look more closely at $\sqrt[3]{u-1} - \sqrt[3]{u+1} = 1$. It seems difficult for the left side to be positive, and indeed we would be done if it is the case that it is non-positive for all real $u$. Indeed, we have that $1 > -1$ and $u+1 > u-1$, hence $\sqrt[3]{u+1} \geq \sqrt[3]{u-1}$ because $\sqrt[3]{x}$ is an increasing function. Now if $u$ were a real solution then $1 = \sqrt[3]{u-1} - \sqrt[3]{u+1} \leq 0$, contradiction.

Answer 4

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

$$\sqrt[3]{(x-3)^2}+\sqrt[3]{(x-3)(x-1)}+\sqrt[3]{(x-1)^2} >\sqrt[3]{|x-3||x-1|}>0$$