Answer
$\{2,4\}$.
Work Step by Step
The given expression is
$\Rightarrow 2(x-1)^{\frac{1}{3}}=(x^2+2x)^{\frac{1}{3}}$
Cube both sides.
$\Rightarrow \left [2(x-1)^{\frac{1}{3}}\right ]^3=\left [ (x^2+2x)^{\frac{1}{3}}\right ]^3$
Multiply exponents on both sides and simplify.
$\Rightarrow 8(x-1)=x^2+2x$
Use the distributive property.
$\Rightarrow 8x-8=x^2+2x$
Add $-8x+8$ to both sides.
$\Rightarrow 8x-8-8x+8=x^2+2x-8x+8$
Simplify.
$\Rightarrow 0=x^2-6x+8$
Rewrite the middle term $-6x$ as $-4x-2x$.
$\Rightarrow 0=x^2-4x-2x+8$
Group the terms.
$\Rightarrow 0=(x^2-4x)+(-2x+8)$
Factor each group.
$\Rightarrow 0=x(x-4)-2(x-4)$
Factor out $(x-4)$.
$\Rightarrow 0=(x-4)(x-2)$
Set each factor equal to zero.
$\Rightarrow x-4=0$ or $x-2=0$
Isolate $x$.
$\Rightarrow x=4$ or $x=2$
The solution set is $\{2,4\}$.
