If n is even number and $\alpha, \beta$ are the roots of the equation $x^2+px +q=0$ and also of the equation $x^{2n} +p^nx^n +q^n =0$......

Question

Question : If n is even number and $\alpha, \beta$ are the roots of the equation $x^2+px +q=0$ and also of the equation $x^{2n} +p^nx^n +q^n =0$ and $f(x) = \frac{(1+x)^n}{1+x^n}$ then $f(\frac{\alpha}{\beta}) = $ ( where $\alpha^n +\beta^n \neq 0), p \neq 0$)

My approach :

$\alpha + \beta = -p, \alpha \beta = q$

let us take $y =x^n$ therefore $x^{2n} +p^nx^n +q^n=0$ will become $y^n+p^ny+q^n=0$

$f(\frac{\alpha}{\beta}) =\frac{(1+\frac{\alpha}{\beta})^n}{1+\frac{\alpha^n}{\beta^n}}= \frac{(\alpha +\beta)^n}{\alpha^n +\beta^n}$ how to proceed further , answer is -1 , please suggest will be of great help.

Answer 1

$\alpha^{2n}+(-(\alpha+\beta))^n\alpha^n+(\alpha\beta)^n=0$

As $n$ is even,

$$\implies \alpha^n(\alpha^n+(\alpha+\beta)^n+\beta^n)=0$$

Assuming $\alpha\ne0,$ $$\dfrac{(\alpha+\beta)^n}{\alpha^n+\beta^n}=-1$$

$$f\left(\dfrac{\alpha}{\beta}\right)=\dfrac{\left(1+\dfrac\alpha\beta\right)^n}{1+\left(\dfrac\alpha\beta\right)^n}=?$$

Answer 2

$\alpha+ \beta =-p$ and $\alpha^n+ \beta ^n=-p^n$ (Sum of roots)

$f(\frac{\alpha}{\beta}) =\frac{(\alpha +\beta)^n}{\alpha^n +\beta^n}=....$

And it's now straightforward.

Answer 3

The condition "$\alpha$ and $\beta$ are the roots" imply: $$(x-\alpha)(x-\beta)=0 \Rightarrow x^2-(\underbrace{\alpha+\beta}_{-p})x+\underbrace{\alpha\beta}_{q}=0\\ (x^n-\alpha^n)(x^n-\beta^n)=0 \Rightarrow x^{2n}-(\underbrace{\alpha^n+\beta^n}_{-p^n})x+\underbrace{\alpha^n\beta^n}_{q^n}=0$$ Hence: $$f(\frac{\alpha}{\beta}) =\frac{(1+\frac{\alpha}{\beta})^n}{1+\frac{\alpha^n}{\beta^n}}= \frac{(\alpha +\beta)^n}{\alpha^n +\beta^n}=\frac{(-p)^n}{-p^n}=\frac{(-p)^{2k}}{-p^{2k}}=\frac{p^{2k}}{-p^{2k}}=-1.$$