Question 944731
Use a substitution,
{{{u=y^(1/3)}}}
{{{u^2=y^(2/3)}}}
So,
{{{u^2-10u+9=0}}}
{{{(u-1)(u-9)=0}}}
.
.
{{{u-1=0}}}
{{{u=1}}}
{{{y^(1/3)=1}}}
{{{y=1}}}
.
.
{{{u-9=0}}}
{{{u=9}}}
{{{y^(1/3)=9}}}
{{{y=9^3}}}
{{{y=729}}}