Let x = .


Then  =  =  = .


Therefore, the given equation   =   takes the form


     = ,  or,  equivalently


     = 0.


The roots of the polynomial   are:

     = ,        (1)

     =     (2)

     =     (3)


Explanation


    This is a polynomial of degree 3. To find zeros for polynomials of degree 3 or higher we use Rational Root Test.


    The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction , 
    where p is a factor of the trailing constant and q is a factor of the leading coefficient.


    The factors of the leading coefficient (8) are 1, 2, 4, 8 . The factor of the constant term (-1) is 1 . 
    Then the Rational Roots Tests yields the following possible solutions:

        ± 1/1, ±1/2, ±1/4, ±1/8 .


    Substitute the POSSIBLE roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.


    If we plug these values into the polynomial P(x), we obtain   = 0.

    To find remaining zeros we use Factor Theorem. This theorem states that if   is root of the polynomial 
    then this polynomial can be divided with qx−p. In this example:


    Divide P(x) with 2x+1

         = .


    Polynomial    can be used to find the remaining roots.


     is a second degree polynomial. To find its roots use the quadratic formula.

Of the found roots (1), (2) and (3),  only positive root (2) can be the value of  x = .  


Therefore,   = .      ANSWER