Question 352103
No, the key to this problem is to recognize that it's a quadratic equation in disguise.
Use a substitution.
Let {{{u=x^(2/3)}}}, then {{{u^2=x^(4/3)}}}, then
{{{x^(4/3)-5x^(2/3)+6=0}}} becomes
{{{u^2-5u+6=0}}}
Then you solve for the solutions to {{{u}}} by factoring, completing the square, or quadratic formula.
Then once you have {{{u}}}, back substitute and find {{{x}}}.