Algebraic Manipulation with Cube Roots Let $a,$ $b,$ $c$ be the real roots of $x^3 - 4x^2 - 32x + 17 = 0.$ Solve for $x$ in $$\sqrt[3]{x - a} + \sqrt[3]{x - b} + \sqrt[3]{x - c} = 0.$$
We probably have to manipulate the $\sqrt[3]{x - a} + \sqrt[3]{x - b} + \sqrt[3]{x - c}$ into something more convenient, so we can actually use it to solve the problem. The first thing I tried was cubing the equation; I quit midway through realizing it was a bad idea (It was really messy.)
Next, I got the stupid idea of trying to actually solve the cubic. I got nowhere. (What I was hoping for were some nice solutions for $x.$)
Now, I'm stuck. Help? Thanks!

Elevating to third power we get
$$
3 \left[x+\left(\sqrt[3]{x-a}+\sqrt[3]{x-b}\right) \left(\sqrt[3]{x-a}+\sqrt[3]{x-c}\right) \left(\sqrt[3]{x-b}+\sqrt[3]{x-c}\right)\right]-(a+b+c)=0
$$
using the original equation gives
$$
3 \left[x-\sqrt[3]{x-a}\sqrt[3]{x-b}\sqrt[3]{x-c}\right]-(a+b+c)=0\\
3 \left[x-\sqrt[3]{(x-a)(x-b)(x-c)}\right]-(a+b+c)=0\\
$$
but
$$
a+b+c=4\\
(x-a)(x-b)(x-c)=x^3-4x^2-32x+17
$$
so the equation becomes
$$
3 \left[x-\sqrt[3]{x^3-4x^2-32x+17}\right]-4=0\\
$$
and isolating the cube root and elevating again to the third power
$$
x^3-4x^2-32x+17=\left(x-\frac{4}{3}\right)^3\\
$$
simplifying we get
$$
1008x=523\quad\implies\quad x_0=\frac{523}{1008}
$$
Be aware that the cube root is intended to be defined on the whole $\mathbb{R},$ otherwise no real solution would exist, given that one of $x_0-a,x_0-b,x_0-c$ is negative.
