Question 515130
I can take a shot at it
{{{ ( 1 - 4x^2 )^(1/6) }}}
{{{ ( 1 - (2x)^2)^(1/6) }}}
I'll make the substitution {{{ z = 2x }}}
{{{ ( 1 - z^2)^(1/6) }}}
{{{ ( 1 + z )^(1/6) * ( 1 - z )^(1/6) }}}
If {{{z}}} is greater than {{{ 1 }}},
or less than {{{-1}}} then one
of these 2 factors is negative. 
I can rewrite as:
{{{( ( 1 + z )^(1/2) * ( 1 - z )^(1/2))^(1/3) }}}
So, any {{{ z }}} greater than {{{1}}} or less than {{{-1}}}
gives me an imaginary factor. If
{{{ z < = 1 }}} or {{{ z > = -1 }}}, the expression is real.
Since {{{ z = 2x }}}
{{{ -1 <= 2x <= 1 }}}
{{{ -1/2 <= x <= 1/2 }}} answer
hope I got it