Answer
$\{1,512\}$.
Work Step by Step
Substitute $x^{\frac{1}{3}} = A$ in the given equation
$\Rightarrow A^2-9A+8=0$
Rewrite the middle term $-9A$ as $-8A-1A$.
$\Rightarrow A^2-8A-1A+8=0$
Group terms.
$\Rightarrow (A^2-8A)+(-1A+8)=0$
Factor each term.
$\Rightarrow A(A-8)-1(A-8)=0$
Factor out $(A-8)$.
$\Rightarrow (A-8)(A-1)=0$
Set each factor equal to zero.
$\Rightarrow A-8=0$ or $A-1=0$
Isolate $A$.
$\Rightarrow A=8$ or $A=1$
Substitute back $A=x^{\frac{1}{3}}$.
$\Rightarrow x^{\frac{1}{3}}=8$ or $x^{\frac{1}{3}}=1$
Cube both sides.
$\Rightarrow x=8^3$ or $x=1^3$
Simplify.
$\Rightarrow x=512$ or $x=1$
The solution set is $\{1,512\}$.
