If α, β the roots of x2+px+q=0 and x2n+pnxn+qn=0 and if (α/β),(β/α) are the roots of xn+1+(x+1)n=0 then n(∈N)

We have α+β=−p and αβ=q             (1)Also, since α, β are the roots of x2n+pnxn+qn=0,We have α2n+pnαn+qn=0 and β2n+pnβn+qn=0Subtracting the above relations, we getα2n−β2n+pnαn−βn=0∴ αn+βn=−pn         (2)Given, α/β or β/α is a root of xn+1+(x+1)n=0. So,       (α/β)n+1+[(α/β)+1]n=0⇒αn+βn+(α+β)n=0⇒−pn+(−p)n=0                  [Using (1) and (2)]It is possible only when n is even.