Question 194506
Let {{{x^(1/3) = y}}}
then the equation becomes
{{{y^2-3y-10=0}}}
Solving for y, we have
(y-5)(y+2)=0
y = 5 or y = -2
From {{{x^(1/3) =5}}}, we get x=125
From {{{x^(1/3) =-2}}}, we get x = -8