Question 572491
I assume that you mean that
{{{y^3-1/8}}} has a zero at 1/2 and that 1/2=0.5
The equations {{{y^3-1/8=0}}}, {{{y^3-1/8+1/8=0+1/8}}} and {{{y^3=1/8}}} are equivalent.
The only solution is {{{y=1/2}}}, because
{{{(1/2)^3=(1/2)(1/2)(1/2)=(1*1*1)/(2*2*2)=1/8}}}
and the only real number that cubed equals {{{1/8}}} is {{{1/2}}}.
From another angle:
In polynomials, one of the special products is difference of cubes. The book may say
{{{a^3-b^3=(a-b)(a^2+ab+b^2)}}}
Applied to {{{y^3-1/8=y^3-(1/2)^3}}},
{{{y^3-1/8=y^3-(1/2)^3=(y-1/2)(y^2+(1/2)y+1/4)=(y-1/2)(y^2+y/2+1/4)}}}