Question 357868
Let {{{u=x^(1/3)}}}
{{{u^2=x^(2/3)}}}
{{{2u^2-3u=20}}}
{{{2u^2-3u-20=0}}}
{{{(2u+5)(u-4)=0}}}
Two solutions:
{{{2u+5=0}}}
{{{2u=-5}}}
{{{u=-5/2}}}
{{{x^(1/3)=-5/2}}}
{{{highlight(x=-125/8)}}}
.
.
.
{{{u-4=0}}}
{{{u=4}}}
{{{x^(1/3)=4}}}
{{{highlight(x=64)}}}