You need to learn Latex my friend. I think I can follow it though.
I see a mistake from the 3rd line to the 4th line when you're FOILing. $\displaystyle x^{\frac{1}{3}} \times x^{\frac{1}{3}} = x^{\frac{2}{3}}$ since you add the exponents in this case. You wrote 1/6 and I think it should be 2/3.
After this I'm losing what you're doing. The right hand side was 2, but now it looks like you're dividing by 4. Where'd that come from?
Anyway, first couple of things I noticed. I'll come back later and give this a more thorough answer if no one has done so. I gotta head out for now.
Hello
As you know now you must to solve
equations system:
x^1/6-5x^1/3+6=0(1)
5x^1/3+10<>0 (2)
in (1) equation x^1/6=-1 no solvation
only x^1/6=6/5
in (2) equation x^1/3=-2 no solvation so 5x^1/3+10 are allways > 0
and we have only one solvation:Ats.: x=46656/15625
Good luck
Let's start from the beginning:
$\displaystyle \frac4{\sqrt[3]{x}+2}+\frac{\sqrt[3]{x}+3}5=2$
Multiply both sides of the equation by $\displaystyle 5(\sqrt[3]{x}+2)$ to get rid of all fractions:
$\displaystyle 20+(\sqrt[3]{x}+3)(\sqrt[3]{x}+2) = 10(\sqrt[3]{x}+2)$
$\displaystyle 20+(\sqrt[3]{x})^2+5 \cdot \sqrt[3]{x} + 6=10\sqrt[3]{x} + 20$
Now substitute $\displaystyle y = \sqrt[3]{x}$:
$\displaystyle y^2-5y+6=0~\iff~y = 2~\vee~y=3$
And therefore
$\displaystyle 2 = \sqrt[3]{x}~\implies~x=8$
$\displaystyle 3 = \sqrt[3]{x}~\implies~x = 27$