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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.

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  • $\begingroup$ Make $x^{1/9}$ your new unknown,say $y$. The equation becomes $r(y^9-1)=y^8-y$. Now the equation is polynomial. $\endgroup$ – OR. Apr 4 '15 at 13:25
  • $\begingroup$ As your tag say, numerical methods only, I bet. If you want me to elaborate, give me an $r$ of your choice. Polynomials of degree $>4$ do not have explicit solutions (in general). And take care that squaring, cubing, ..., introduce extra roots. You have the trivial $x=1$ as solution already. $\endgroup$ – Claude Leibovici Apr 4 '15 at 13:26
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    $\begingroup$ Better. The substitution $y=x^{1/9}+\frac{1}{x^{1/9}}$ reduces it into a degree $4$ polynomial for which we have formulas to compute the roots. $\endgroup$ – OR. Apr 4 '15 at 13:34
  • $\begingroup$ You should look at Galois group of the equation wrote by @Mlazhinka Shung Gronzalez LeWy to see if it's soluble via radicals only $\endgroup$ – Renato Faraone Apr 4 '15 at 13:37
  • $\begingroup$ @RenatoFaraone The equation is of course solvable in radicals. No need to look at Galois groups. $\endgroup$ – OR. Apr 4 '15 at 13:39
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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.

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  • $\begingroup$ I guess that tedious is an understatement ! Thanks for the nice answer. $\endgroup$ – Claude Leibovici Apr 4 '15 at 14:01
  • $\begingroup$ I like this answer, but I think the final expansion is not quite right. With $z=y+\frac 1y$, I get $y^4+\frac 1{y^4}=z^4-4z^2+2, y^3+\frac 1{y^3}=z^3-3z, y^2+\frac 1{y^2}=z^2-2$ - I think you've taken $-4z^3$ from the first term, when it should be $-4z^2$ - the powers in each of these expansions are either all even or all odd. $\endgroup$ – Mark Bennet Apr 4 '15 at 17:06
  • $\begingroup$ @MarkBennet Yes, I wrote it in the air, so probably many subtractions are wrong. $\endgroup$ – OR. Apr 4 '15 at 17:11
  • $\begingroup$ If anyone are interested in knowing what the solutions in radicals are, look no further. $\endgroup$ – Alice Ryhl Apr 4 '15 at 17:15
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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 $y=1$

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  • $\begingroup$ Yes I'm sorry I did not pay attention $\endgroup$ – Elaqqad Apr 4 '15 at 13:44

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