If one root is the n^th power of the other root of the equation x² –ax + b = 0, then ${\mathrm{b}}^{\frac{\mathrm{n}}{\mathrm{n}+1}}+{\mathrm{b}}^{\frac{1}{\mathrm{n}+1}}=$

The given equation is $x^2-ax+b=0$

Given that for the above equation, one root is n times the other

suppose that $\alpha ,{\alpha }^n$ are the roots of the above equation

Sum of the roots is $\alpha +{\alpha }^n=a$ and product of the roots is ${\alpha }^{n+1}=b\Rightarrow \alpha =b^{\frac{1}{n+1}}$

Eliminate $\alpha$ from the above two equations to get the required condition 

Hence,  $$\begin{gathered} \mathrm{\alpha },{\mathrm{\alpha }}^{\mathrm{n}}=\mathrm{b};\mathrm{\alpha }+{\mathrm{\alpha }}^{\mathrm{n}}=a \\ {b}^{1/\mathrm{n}+1}+{\mathrm{b}}^{\mathrm{n}/\mathrm{n}+1}=\mathrm{a} \end{gathered}$$