Solve $\sqrt[3]{1+x}+\sqrt[3]{1-x}=\sqrt[3]{2}$

Question

Solve $\sqrt[3]{1+x}+\sqrt[3]{1-x}=\sqrt[3]{2}$

$\Rightarrow(\sqrt[3]{1+x}+\sqrt[3]{1-x})^3=(\sqrt[3]{2})^3$

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

$1+x+3[(1+x)^{\frac{2}{3}}(1-x)^{\frac{1}{3}}+(1+x)^{\frac{1}{3}}(1-x)^{\frac{2}{3}}]+1-x=2$

$3[(1+x)^{\frac{2}{3}}(1-x)^{\frac{1}{3}}+(1+x)^{\frac{1}{3}}(1-x)^{\frac{2}{3}}]=0$

$(1+x)^{\frac{2}{3}}(1-x)^{\frac{1}{3}}+(1+x)^{\frac{1}{3}}(1-x)^{\frac{2}{3}}=0$

$(1+x)^{\frac{1}{3}}(1-x)^{\frac{1}{3}}[(1+x)^{\frac{1}{3}}+(1-x)^{\frac{1}{3}}]=0$

So $x=\pm1$, which is the answer.

My question is about the remaining part $[(1+x)^{\frac{1}{3}}+(1-x)^{\frac{1}{3}}]=0$.

$\Rightarrow[(1+x)^{\frac{1}{3}}]^3=[-(1-x)^{\frac{1}{3}}]^3$

$1+x=-(1-x) $

$1+x=-1+x$

$1=-1$

So does this mean my assumption $[(1+x)^{\frac{1}{3}}+(1-x)^{\frac{1}{3}}]=0$ is wrong and it is not zero? Then how would I find its value? Thank you.

Answer 1

You are starting from the assumption that $x$ is a real number such that $$\sqrt[3]{1+x}+\sqrt[3]{1-x}=\sqrt[3]{2}.\tag{1}$$ You then arrive at three cases; either $x=\pm1$ or $$(1+x)^{\frac{1}{3}}+(1-x)^{\frac{1}{3}}=0.$$ But this last case contradicts the original assumption $(1)$ directly. So there is no other solution.

Answer 2

My question is about the remaining part $[(1+x)^{\frac{1}{3}}+(1-x)^{\frac{1}{3}}]=0$.

Because you force it equals $0$, so you get contradiction. Actually, if we don't include complex numbers, within real number domain, we have

$$(1+x)^{\frac{1}{3}}+(1-x)^{\frac{1}{3}}>0,~~\forall x$$

Here is the plot

Which means this term is always postive. It is like when you deal with $x^2-x+1$, you treat this term as always positive term.

So go back to this step:

$(1+x)^{\frac{1}{3}}(1-x)^{\frac{1}{3}}[(1+x)^{\frac{1}{3}}+(1-x)^{\frac{1}{3}}]=0$

Since this term is always positive, we can safely cancel it, and get

$$(1+x)^{\frac{1}{3}}(1-x)^{\frac{1}{3}}=0$$

and you get two roots:

$$x=1,~~x=-1$$

Answer 3

As noticed, the supplemental solutions you are looking for are not compatible with the original equation, indeed their are complex and we are not interested with them.

As an alternative to avoid additional equations, we can let $1+x=z^3 \implies 1-x=2-z^3$ and then

$$\sqrt[3]{1+x}+\sqrt[3]{1-x}=\sqrt[3]{2} \iff 3\sqrt[3]{2}z^2-3\sqrt[3]{2^2}z=0$$

which leads to $z=0$ or $z= \sqrt[3]{2} $ that is $x=\pm 1$.