Question 355271
Let {{{u=x^(1/3)}}}, {{{u^2=x^(2/3)}}}
{{{u^2-u-12=0}}}
{{{(u-4)(u+3)=0}}}
Two solutions:
{{{u-4=0}}}
{{{u=4}}}
{{{x^(1/3)=4}}}
{{{highlight(x=64)}}}
.
.
{{{u+3=0}}}
{{{u=-3}}}
{{{x^(1/3)=-3}}}
{{{highlight(x=-27)}}}