R(1)

If $\displaystyle \alpha$ and $\displaystyle \beta$ are the roots of the equations $\displaystyle {{x}^{2}}+px+q=0$ and $\displaystyle {{x}^{{2n}}}+{{p}^{n}}{{x}^{n}}+{{q}^{n}}=0$ , where $\displaystyle n$ is an even integer . Prove that $\displaystyle \frac{\alpha }{\beta }$ and $\displaystyle \frac{\beta }{\alpha }$ are the roots of the equation $\displaystyle {{x}^{n}}+1+{{(x+1)}^{n}}=0$ .

Solution

$\displaystyle {{x}^{2}}+px+q=0$

$\displaystyle \alpha \ \ \text{and  }\beta \ \ \ \text{are the roots of the equation }\text{.}$

$\displaystyle p=-(\alpha +\beta )\ \ \text{and   }q=\alpha \beta$

$\displaystyle {{x}^{{2n}}}+{{p}^{n}}{{x}^{n}}+{{q}^{n}}=0$

$\displaystyle \alpha \ \ \text{and  }\beta \ \ \ \text{are the roots of the equation }\text{.}$

$\displaystyle {{\alpha }^{{2n}}}+{{p}^{n}}{{\alpha }^{n}}+{{q}^{n}}=0\ \ \ \text{and   }{{\beta }^{{2n}}}+{{p}^{n}}{{\beta }^{n}}+{{q}^{n}}=0\text{ }$

$\displaystyle {{\alpha }^{{2n}}}+{{p}^{n}}{{\alpha }^{n}}={{\beta }^{{2n}}}+{{p}^{n}}{{\beta }^{n}}$

$\displaystyle {{\alpha }^{{2n}}}-{{\beta }^{{2n}}}={{p}^{n}}({{\beta }^{n}}-{{\alpha }^{n}})$

$\displaystyle ({{\alpha }^{n}}-{{\beta }^{n}})({{\alpha }^{n}}+{{\beta }^{n}})={{(\alpha +\beta )}^{n}}({{\beta }^{n}}-{{\alpha }^{n}})\ \left[ {\because n\ \text{is even integer}} \right]$

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

$\displaystyle \text{Let  }f(x)={{x}^{n}}+1+{{(x+1)}^{n}}$

$\displaystyle \ \ \ \ \ f(\frac{\alpha }{\beta })={{(\frac{\alpha }{\beta })}^{n}}+1+{{(\frac{\alpha }{\beta }+1)}^{n}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{\alpha }^{n}}+{{\beta }^{n}}+{{{(\alpha +\beta )}}^{n}}}}{{{{\beta }^{n}}}}=0$

$\displaystyle \ \ \ \ \ f(\frac{\beta }{\alpha })={{(\frac{\beta }{\alpha })}^{n}}+1+{{(\frac{\beta }{\alpha }+1)}^{n}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{\beta }^{n}}+{{\alpha }^{n}}+{{{(\alpha +\beta )}}^{n}}}}{{{{\alpha }^{n}}}}=0$

$\displaystyle \therefore \frac{\alpha }{\beta }\ \ \text{and  }\frac{\beta }{\alpha }\ \ \text{are the roots of the equation }{{x}^{n}}+1+{{(x+1)}^{n}}=0.$