Question 96205
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Please help me solve this problem:

y^4/3 = -3y

This is what I have done:
y^4/3 + 3y =0

1. The exp of y is understood to be 1...so does this mean that I need to find the commoon
denominator of the exp so that I could add?  Should the exp of 3y then be 1/3?
y^4/3 + 3y^1/3= 0

2. Then in the book it says to factor out maybe 1/3?

This is where I get confused, because the problem in the book that is simmilar looks like this:
x^3/2= x^1/2
x^3/2- x^1/2= 0
x^1/2(x-1)) =0   How are they factoring out x^1/2?  Is x^1/2(x-1) = x^3/2 -x^1/2

3. So once they get x^1/2= 0   and x-1=0

so the answer is x=0 and x=1 

I think you're CONFUSING yourself, and probably mixing up this given problem with the one in your book.
Anyway, I hope you can follow what I've presented below. 

            {{{matrix(2,1, " ", y^(4/3) = - 3y)}}}
        {{{matrix(2,1, " ", y^(4/3)) + 3y = 0}}} ---- Adding {{{3y^2}}} to both sides
     {{{matrix(2,1, " ", y(y^(1/3) + 3) = 0)}}} ---- Factoring out y
         {{{matrix(2,1, " ", y^(1/3)) + 3 = 0}}}    OR    <font color = red><font size = 4><b> y = 0</font></font></b>  
           {{{matrix(2,1, " ", y^(1/3)) = - 3}}}
       {{{matrix(2,1, " ", y^((1/3)^3)) = (- 3)^3}}} --- Cubing each side
              {{{highlight(y) = (- 3)^3 = highlight(- 27))}}}</pre>