If $ax+by=7$, . 
If 
  $$ax+by=7$$
  $$ax^2+by^2=49$$
  $$ax^3+by^3=133$$
  $$ax^4+by^4=406$$
  then find the value of
  $$2014(x+y-xy) - 100(a+b)$$

My attempt:
$$ax^2+by^2=49$$
$$ax^2+by^2=(ax+by)^2$$
$$ax^2+by^2=a^2x^2+2abxy+b^2y^2$$
$$ax^2-a^2x^2+by^2-b^2y^2=2abxy$$
$$ax^2(1-a)+by^2(1-b)=2abxy$$

Hint $ $ Use $\ ax^{n+1}\!+by^{n+1} = (x\!+\!y)(ax^n\!+by^n) - xy(ax^{n-1}\!+by^{n-1})$ to solve for $\,xy,x\!+\!y$
Remark $ $ The recurrence is $\,(S-x)(S-y)f_n = (S^2 - (x+y)S+xy)f_n = 0\,$ where $\,Sf_n = f_{n+1}.\,$ It has solutions $\,f_n = x^n,\,y^n\,$ so also $\,f_n = ax^n + bx^n$ by linearity.
