Question 947118
Wait, check again.
{{{3(1)^(1/3)+2(1)^(2/3)=5}}}
{{{3+2=5}}}
{{{5=5}}}
It does work.
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This is actually a quadratic equation.
Use a substitution.
{{{u=x^(1/3)}}}
{{{u^2=x^(2/3)}}}
So then,
{{{3u+2u^2=5}}}
{{{2u^2+3u-5=0}}}
{{{(u-1)(2u+5)=0}}}
Two "u" solutions:
{{{u-1=0}}}
{{{u=1}}}
{{{x^(1/3)=1}}}
{{{x=1}}}
and
{{{2u+5=0}}}
{{{2u=-5}}}
{{{u=-5/2}}}
{{{x^(1/3)=-5/2}}}
{{{x=-125/8}}}