Solving equation with rational exponent I have this equation: $$\mathrm{r} (x-1) = x^{8/9} - x^{1/9}$$
where $\mathrm{r}$ is a constant. Is there a general technique to solve such equations? Raising it to the 9nth power: $$\mathrm{r}^9 (x-1)^9 = (x^{8/9} - x^{1/9})^{9} \\ \mathrm{r}^9 (x-1)^9  = x (x^{7/9} - 1)^9\\$$ seems quite cumbersome.
 A: First make the substitution $y=x^{1/9}$ the equation becomes
$$r(y^9-1)=y^8-y$$
We can factor out from both sides $y-1$, giving us one solution $y=1$, and therefore the solution $x=1$. What remains is
$$r(y^8+y^7+y^6+y^5+y^4+y^3+y^2+y+1)=y(y^6+y^5+y^4+y^3+y^2+y+1)$$
or
$$ry^8+(r-1)\sum_{k=1}^{7}y^k+r=0$$
Observe that the coefficients of this polynomial are symmetric. Therefore we can use the substitution $z=y+\frac{1}{y}$. Observe that if we divide the polynomial above by $y^4$ we get 
$$r\left(y^4+\frac{1}{y^4}\right)+(r-1)\sum_{k=1}^{3}\left(y^k+\frac{1}{y^k}\right)+(r-1)=0$$
Each $y^k+\frac{1}{y^k}$ can be written as a linear combination of powers of $\left(y+\frac{1}{y}\right)^n$, $n=0,1,2,...,k$. Therefore this can be written as a degree $4$ polynomial in $z=y+\frac{1}{y}$.
Tedious! But well... we get (using computations by Mark Bennet)
$$r(z^4-4z^2+2)+(r-1)(z^3-3z)+(r-1)(z^2-2)+(r-1)z+(r-1)=0$$
or
$$rz^4+(r-1)z^3-(3r+1)z^2+(2-2r)z+(r+1)=0$$
But check it yourself just in case.
Now, the degree $4$ polynomial equations we can solve in radicals.
A: To solve this equation you can take :$y=x^{\frac{1}{9}}$ and then:
$$r(y^9-1)=y^8-y$$
so and here you have a polynomial equation, if you're lucky the equation will have degree less than $4$ or you can reduce it by finding some intuitive solutions, otherwise there is no a general form for this type of equations.
For this equation there is two obvious solutions $y=1$
