3x^2/3 + x^1/3 - 2 = 0

how to go about this ?
no calculator question !

This is of the form 3y² + y - 2 = 0 where y = x^1/3

=> (3y - 2)(y + 1) = 0

so, y = 2/3 or y = -1

Now, x = y³

=> x = (-1)³ = -1 or x = 8/27

:)>

Don't look at the fraction exponents for now. Treat it like 3x^2+x-2=0. You will have to use the 1/3 in a minute, but for now that complicates things. I just means you will take the cubed root at one point.
So fact that
(3x-2)(x+1)
Then the x values are to the 1/3 exponent
3x^(1/3)-2=0
x^(1/3)+1=0
then x=(2/3)^(1/3) and x=1

3x^2/3 + x^1/3 - 2 = 0

(3x^2/3 + x^1/3 - 2 = 0)^3 -------> 9x^2 + x^3 -8 =0

9x^2 + x^3 = 8

x^2 (9+x) = 8

Now you can do a little bit of guessing, since we know that 8 = either 1*8 or 2*4.
It turns out that we can't have 2*4 = 8, so the equation should represent 1*8 = 8.
Therefore, by plugging in x=-1, you will get (-1)^2 (9+(-1)) = 8
...and there's your answer.

Let us define a new variable u=x^1/3.
This equation in terms of u is 3u^2+u-2=0.
This one is a quadratic one who has even an obvious root-1.Since the product is -2/3, the other root is 2/3.
So we have two solutions for u, -1 and 2/3 which we need to translate into x.
From our definition, x=u^3 so the solutions in terms of x are (-1)^3 and (2/3)^3.
So we have -1 and 8/27 as the two roots of the original equation.

do you mean solve for x?
or differentiate?
or what?
specify