Sum of real roots

$\begin{aligned} \sqrt[3]{1-x} + \sqrt[3]{x+3} & = \frac 52 \\ \left(\sqrt[3]{1-x} + \sqrt[3]{x+3} \right)^3 & = \left(\frac 52\right)^3 \\ 1-x + 3\sqrt[3]{(1-x)^2(x+3)} + 3\sqrt[3]{(1-x)(x+3)^2} + x + 3 & = \frac {125}8 \\ 4 + 3\sqrt[3]{(1-x)(x+3)}\left({\color{#3D99F6}\sqrt[3]{1-x} + \sqrt[3]{x+3}} \right) & = \frac {125}8 \\ 3\left({\color{#3D99F6}\frac 52} \right) \sqrt[3]{(1-x)(x+3)} & = \frac {125}8 - 4 \\ \sqrt[3]{(1-x)(x+3)} & = \frac {31}{20} \\ 3 -2x - x^2 & = \left(\frac {31}{20}\right)^3 \\ \implies x^2+{\color{#D61F06}2}x - 3 + \left(\frac {31}{20}\right)^3 & = 0 \end{aligned}$

Since $2^2 - 4 \left(-3 + \left(\frac {31}{20}\right)^3 \right) > 0$ , there are two real roots $x$ . By Vieta's formula , we have sum of the two real roots equal to $\boxed{{\color{#D61F06}-2}}$ .

Method 1

Let $a=\sqrt [ 3 ]{ 1-x }$ and $b=\sqrt [ 3 ]{ x+3 }$ . Now $a+b=\frac{5}{2}$ . If each side is raised to the 3rd power, we have $a^3+b^3+3ab(a+b)=\frac{125}{8}$ . Substitute $a+b=\frac{5}{2}$ and $a^3+b^3 = 4$ to the left-hand-side, and finally obtain $ab=\frac{31}{20}$ .

Since $a+b=\frac{5}{2}$ and $ab=\frac{31}{20}$ , we note that $a$ and $b$ are roots to the equation $u^2-\frac{5}{2}u+\frac{31}{20}=0$ . Solve this quadratic equation and it gives $u=\frac{5}{4} \pm \frac{1}{4\sqrt{5}}$ .

This means that $a=\sqrt [ 3 ]{ 1-x } =\frac{5}{4} + \frac{1}{4\sqrt{5}}$ or $a=\sqrt [ 3 ]{ 1-x } =\frac{5}{4} - \frac{1}{4\sqrt{5}}$ . Hence $x=1-\left(\frac{5}{4} + \frac{1}{4\sqrt{5}}\right)^3$ or $x=1-\left(\frac{5}{4} - \frac{1}{4\sqrt{5}}\right)^3$ . Note that each of these is a real number.

The sum of real roots is $2-\left[\left(\frac{5}{4} + \frac{1}{4\sqrt{5}}\right)^3+\left(\frac{5}{4} - \frac{1}{4\sqrt{5}}\right)^3\right] = 2-2\left[ \left(\frac{5}{4}\right)^3+3\left(\frac{5}{4}\right)\left(\frac{1}{4\sqrt{5}}\right)^2\right] = 2-2\left[ \frac{125}{64}+\frac{3}{64}\right]=\boxed{-2}$ .

Method 2

Since $\sqrt [ 3 ]{ 1-x }+\sqrt [ 3 ]{ x+3 }=\sqrt [ 3 ]{ 2-(x-1) }+\sqrt [ 3 ]{ 2+(1-x) }$ . This means that the graph of $f(x)=\sqrt [ 3 ]{ 1-x }+\sqrt [ 3 ]{ x+3 }$ is symmetry about $x=-1$ . By sketching the graph of $y=f(x)$ , we know that there are two solutions to $f(x)=\frac{5}{2}$ . Then the solution is of $x_1=-1+\alpha$ and $x_2=-1-\alpha$ for some $\alpha>0$ . So the sum of roots is $\boxed{-2}$ .