Answer
$\{0,10\}$.
Work Step by Step
The given expression is
$\Rightarrow 3x^{\frac{1}{3}}=(x^2+17x)^{\frac{1}{3}}$
Cube both sides.
$\Rightarrow \left [3x^{\frac{1}{3}}\right ]^3=\left [ (x^2+17x)^{\frac{1}{3}}\right ]^3$
Multiply exponents on both sides and simplify.
$\Rightarrow 27x=x^2+17x$
Subtract $27x$ from both sides.
$\Rightarrow 27x-27x=x^2+17x-27x$
Simplify.
$\Rightarrow 0=x^2-10x$
Factor out $x$.
$\Rightarrow 0=x(x-10)$
Set each factor equal to zero.
$\Rightarrow x=0$ or $x-10=0$
Isolate $x$.
$\Rightarrow x=0$ or $x=10$
The solution set is $\{0,10\}$.
