If alpha and beta αandβ are the roots of the equation x^2-p(x+1)-c=0x2−p(x+1)−c=0 or
x^2-px-(p+c)=0x2−px−(p+c)=0
then
alpha+beta=pand alphabeta=-(p+c)α+β=pandαβ=−(p+c)
So c=-alpha-beta-alphabetac=−α−β−αβ
Now the numerical value of (alpha^2+2alpha+1)/(alpha^2+2alpha+c)+(beta^2+2beta+1)/(beta^2+2beta+c)α2+2α+1α2+2α+c+β2+2β+1β2+2β+c
=(alpha^2+2alpha+1)/(alpha^2+2alpha-alpha-beta-alphabeta)+(beta^2+2beta+1)/(beta^2+2beta-alpha-beta-alphabeta)=α2+2α+1α2+2α−α−β−αβ+β2+2β+1β2+2β−α−β−αβ
=(alpha^2+2alpha+1)/(alpha^2+alpha-beta-alphabeta)+(beta^2+2beta+1)/(beta^2+beta-alpha-alphabeta)=α2+2α+1α2+α−β−αβ+β2+2β+1β2+β−α−αβ
=(alpha+1)^2/(alpha(alpha+1)-beta(alpha+1))+(beta+1)^2/(beta(beta+1)-alpha(1+beta))=(α+1)2α(α+1)−β(α+1)+(β+1)2β(β+1)−α(1+β)
=(alpha+1)^2/((alpha-beta)(alpha+1))-(beta+1)^2/((beta+1)(alpha-beta))=(α+1)2(α−β)(α+1)−(β+1)2(β+1)(α−β)
=(alpha+1)/(alpha-beta)-(beta+1)/(alpha-beta)=α+1α−β−β+1α−β
=(alpha+1-beta-1)/(alpha-beta)=α+1−β−1α−β
=1=1