Take the 6th. power of all terms
(x^5/6)^6 + (x^2/3)^6-(2x^1/2)^6 =0 Indices (a^m)^n= A^mn
x^5+x^4-2x^3=0
x^3(x^2+x-2)=0
x^3(x^2+2x-x-2)=0
x^3{(x(x+2)-1(x+2)}=0
x^3{(x-1)(x+2)=0
x^3(x-1)(x+2)=0
The solution in the post by @mananth is FATALLY incorrect.
His first step is to raise all three addends in the left side of the equation to degree 6.
But this step is not an equivalent transformation, so, starting from this point,
his solution is inadequate.
See my correct solution below.
The original equation is
x^5/6 + x^2/3 - 2x^1/2 = 0. (1)
It is the same as
x^5/6 + x^4/6 - 2x^3/6 = 0. (2)
Factor left side, taking x^3/6 out the parentheses as a common factor
x^3/6*(x^2/6 + x^1/6 - 2) = 0. (3)
One root is x = 0, generated by the common factor x^3/6.
To find the roots generated by the expression in parentheses, introduce new variable t = x^1/6.
Then the expression in parentheses is
t^2 + r - 2.
Its roots are = = = = .
So, the roots are t = -2 and t = 1.
Since t = x^1/6, we accept the positive root t = 1 and reject the negative root t = -2.
Thus, finally we have the solutions to the original equation x = 0 and x = 1.
ANSWER. The original equation has two real solutions x = 0 and x = 1.
