Question 358033
Substitute, let {{{u=x^(1/3)}}}
{{{u^2=x^(2/3)}}}
{{{u^2+7u+12=0}}}
{{{(u+4)(u+3)=0}}}
Two solutions:
{{{u+4=0}}}
{{{u=-4}}}
{{{x^(1/3)=-4}}}
{{{highlight(x=-64)}}}
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.
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{{{u+3=0}}}
{{{u=-3}}}
{{{x^(1/3)=-3}}}
{{{highlight(x=-27)}}}