Assuming that the correct formulation is
If #x/a+y/b=1 and x^2/a+y^2/b=(ab)/(a+b)#
then #x^n+1/a+y^n+1/b=#
From
#{(x/a+y/b=1),(x^2/a+y^2/b=(ab)/(a+b)):}#
solving for #x,y# we obtain
#x=y=(ab)/(a+b)#
To obtain this result we solve for #X=x/a,Y=y/b#
#{(X+Y=1),(aX^2+bY^2=(ab)/(a+b)):}#
#aX^2+b(1-X)^2=(ab)/(a+b)# and solving for #X#
#(a+b)X^2-2bX+b=(ab)/(a+b)# or
#X^2-(2b)/(a+b)X+(b/(a+b))^2# or
#(X-b/(a+b))^2=0#
then #X=b/(a+b)# and analogously #Y=a/(a+b)#
so #x=y=(ab)/(a+b)#
Finally we have
#x^n+1/a+y^n+1/b=2((ab)/(a+b))^n+1/a+1/b#
