Question 698364
You probably meant x^(2/3)-6x^(1/3)+5=0
With the change of variable y=x^(1/3) --> x^(2/3)=[x^(1/3)]^2=y^2 ,
the equation turns into the quadratic equation
{{{y^2-6y+5=0}}} with solutions {{{y=1}}} and {{{y=5}}}
 
Going back to {{{x}}} ,
y=x^(1/3) <--> {{{y^3=x}}} , so
{{{y=1}}} --> {{{y^3=x=1^3=1}}} --> {{{highlight(x=1)}}}
{{{y=5}}} --> {{{y^3=x=5^3=125}}} --> {{{highlight(x=125)}}}