1. ## stuck....

ok I got this far, everything seems to be nice and neat, no crazy numbers, but what do I do now??

2. 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. $x^{\frac{1}{3}} \times x^{\frac{1}{3}} = x^{\frac{2}{3}}$ 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.

3. 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

4. oh i`m so sorry, because i looked only in the last line

5. ok made some more progress. Now though I have what seems like a quadratic equation but it is made of fractions? what to do??

6. Originally Posted by yuriythebest
ok made some more progress. Now though I have what seems like a quadratic equation but it is made of fractions? what to do??
Susbstitute $u = x^{\frac13}$
Now the last equation looks like this:
$u^2 - 5u + 22 = 0$
You will not be very happy. It has complex roots I think

7. Originally Posted by yuriythebest
ok I got this far, everything seems to be nice and neat, no crazy numbers, but what do I do now??
Let's start from the beginning:

$\frac4{\sqrt[3]{x}+2}+\frac{\sqrt[3]{x}+3}5=2$

Multiply both sides of the equation by $5(\sqrt[3]{x}+2)$ to get rid of all fractions:

$20+(\sqrt[3]{x}+3)(\sqrt[3]{x}+2) = 10(\sqrt[3]{x}+2)$

$20+(\sqrt[3]{x})^2+5 \cdot \sqrt[3]{x} + 6=10\sqrt[3]{x} + 20$

Now substitute $y = \sqrt[3]{x}$:

$y^2-5y+6=0~\iff~y = 2~\vee~y=3$

And therefore
$2 = \sqrt[3]{x}~\implies~x=8$
$3 = \sqrt[3]{x}~\implies~x = 27$

8. ok thanks!