Finding roots of equation $$r\left(\frac 1p -1\right)- \left(\frac 1p\right)^\frac89 + \left(\frac 1p\right)^\frac19 =0$$
where $ r = 2^\frac89 - 2^\frac19$
How do I solve this without a computer?
 A: Since $r=2^{8/9}-2^{1/9}$, it seems like it might be possible to choose $\frac{1}{p}=2$, i.e., $p=\frac{1}{2}$.  This makes the expression become $r(2-1)-2^{8/9}+2^{1/9}=r-r=0$.  Therefore $p=\frac{1}{2}$ is a root.
If $\frac{1}{p}=1$, then the first term vanishes and $1^{8/9}=1$ and $1^{1/9}=1$.  Therefore $1$ is another real root.
As @user44394 states, observe that $\frac{1}{2}(2^{8/9}-2^{1/9})=2^{-1/9}-2^{-8/9}=(\frac{1}{2})^{8/9}-(\frac{1}{2})^{1/9}$ giving the third (and last by Descartes rule of signs) root of $p=2$.
A: First of all, there no need to express as 1/p so
$(2^{\frac 8 9} - 2^\frac 1 9)(m - 1) - m^{\frac 8 9} + m^{\frac 1 9} = 0$
$(2^{\frac 8 9} - 2^\frac 1 9)(m - 1) - (m^{\frac 8 9} - m^{\frac 1 9}) = 0$
Oh look! The $(m^{\frac 8 9} - m^{\frac 1 9})$ looks like  $(2^{\frac 8 9} - 2^\frac 1 9)$!  What if $m = 2$?  Then $m - 1$ = 1 and we'd get an equality.  So $p = \frac 1 2$ is a solution.
But that was us seeing something.  Ho do we solve in general?
$(2^{\frac 8 9} - 2^\frac 1 9) = \frac{ (m^{\frac 8 9} - m^{\frac 1 9})}{m-1} $
$2^{\frac 1 9}(2^8 - 2) = \dfrac {m'^8 - m} {m'^9 - 1}$ for $m' = m^{\frac 1 9}$....
You know what? let's just say $p = \frac 1 2$ and call it a day.
