# Find an algebraic function

1. Jul 1, 2017

### Karol

1. The problem statement, all variables and given/known data

2. Relevant equations
Algebraic Fuction:
$$P_0(x)y^n+P_1(X)y^{n-1}+...+P_{n-1}(x)y+P_n(x)=0$$
$P_i(x)~$ are polinomials, $~P_0(x)\neq 0$

3. The attempt at a solution
$$\left( x^{2/3} \right)^3 (1-x)+\left( x^{2/3} \right)^2(x)=x^2-x^3+x^{5/3}$$
Any integer exponent can't cancel the $~x^{5/3}$

2. Jul 1, 2017

### Staff: Mentor

I assume, here is $y=f(x)$. So what is $y^3$?

3. Jul 1, 2017

### Karol

$$y=x^{2/3} \rightarrow~y^3=\left( x^{2/3} \right)^3=x^2$$

4. Jul 1, 2017

### Staff: Mentor

Yes, and now you need to write it $...=0$ to see what $P_0(x)$ and $P_3(x)$ are.

5. Jul 1, 2017

### Karol

$$P_0(x)x^2+x^{\frac{5}{3}}+P_3(x)=0$$
P0 and P3 are dependent on each other, i have to choose one:
$$\Rightarrow~P_0(x)=-\frac{-P_3(x)-x^{5/3}}{x^2}-x^{-\frac{1}{3}}$$
No polinomial will result

6. Jul 1, 2017

### Staff: Mentor

Why so complicated? It is far, far easier than this. You already have $y^3-x^2= 0$. All you have to do is to write down the coefficients of $y^3,y^2,y^1,y^0$.

7. Jul 1, 2017

### Karol

I don't have $~y^3-x^2= 0$. I assume you use only: $~\left( x^{2/3} \right)^3 (1-x)=x^2-x^3$, or, maybe you have made a mistake, it's: $~\left( x^{2/3} \right)^3 (1-x)=y^3-x^3$
The coefficient of y3 is, as i chose: $~(1-x)$
$$(1-x)y^3+P_1(x)y^2+P_2(x)y-x^3=0$$
Is it the equation? i arbitrarily chose 3d degree and the coefficient P0=1-x

Last edited: Jul 1, 2017
8. Jul 1, 2017

### LCKurtz

Why did you choose $1-x$? Why not just $1$?

9. Jul 1, 2017

### Staff: Mentor

You only have to compare $y^3-x^2= 0$ with $P_0(x)y^3+P_1(x)y^2+P_2(x)y+P_3(x)=0$.
Now what are the choices for the $P_i$?
$$1\cdot y^3+0\cdot y^2+0\cdot y-x^2=0$$