Cube rooted equation

In the hope of simplifying the given equation we rewrite the given terms as, $\sqrt[3]{x+1}=a$ $\sqrt[3]{x-1}=b$ $\;\;\;\; \implies 3x-1=2b^3+a^3$ From the given equation, we have $(\sqrt[3]{3x-1})^3=(\sqrt[3]{x+1}-\sqrt[3]{x-1})^3$ $\;\;\;\;\;\;\;\;\;\;\;\;\implies 2b^3+a^3=(a-b)^3=a^3-3ab(a-b)-b^3$ This gives us the relation $b^3=-ab(a-b)$ which leads us to our first solution, $b=0\implies x=1$ If $b\neq 0$ , we have $b^2=-a(a-b)\implies a^2+b^2=ab$ Since $a^3=x+1$ and $b^3=x-1$ , we observe that $a^3-b^3=2\implies (a-b)(a^2+ab+b^2)=2$ Using the previous relations, $(a-b)(a^2+ab+b^2)=(a-b)(2ab)=-2b^3=2$ which leads us to the solution $b=-1\implies x=0$ But $x=0$ does not satisfy the original equation.

Therefore, $\boxed{x=1}$ is the only solution to the given equation

$\begin{aligned} \sqrt[3]{x-1} + \sqrt[3]{3x-1} & = \sqrt[3]{x+1} \\ \sqrt[3]{3x-1} & = \sqrt[3]{x+1} - \sqrt[3]{x-1} & \small \blue{\text{Cubing both sides}} \\ 3x - 1 & = (x+1) - 3\sqrt[3]{(x+1)^2(x-1)} + 3\sqrt[3]{(x+1)(x-1)^2} - (x-1) \\ 3x - 1 & = 3\sqrt[3]{x^2-1}\left(\sqrt[3]{x-1} - \sqrt[3]{x+1} \right) + 2 \\ 3x - 3 & = 3\sqrt[3]{x^2-1}\left(-\sqrt[3]{3x-1}\right) & \small \blue{\text{Divide both sides by 3}} \\ x-1 & = - \sqrt[3]{(x^2-1)(3x-1)} & \small \blue{\text{Cubing both sides}} \\ x^3-3x^2+3x-1 & = - 3x^3 + x^2 + 3x - 1 \\ 4x^3 - 4x^2 & = 0 & \small \blue{\text{For }x \ne 0} \\ \implies x & = 1 \end{aligned}$

The sum of all real solutions is $\boxed 1$ .

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

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

So $4x^2(x-1)=0$

Real solution is ${1 }$ . Sum of the solutions is $\boxed{1}$