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?

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 $p=2$.
