Finding the value of $ax^4+by^4$ 
If $\quad a+b=23 , \quad ax+by=79,\quad ax^2+by^2=217,\quad
 ax^3+by^3=691\quad$ find the value of $ax^4+by^4$.

Here is my attempt:
$$(a+b)(x+y)=ax+by+ay+bx\rightarrow 23(x+y)=79+(ay+bx)$$
$$(ax+by)(x+y)=ax^2+by^2+axy+bxy\rightarrow79(x+y)=217+23 xy$$
In each equation I have two unknowns it seems that doesn't work.

Your method is not in vain. We have:
$$(ax+by)(x+y) = ax^2 + by^2 +axy + bxy \to 79(x+y) = 217 + 23xy$$
$$(ax^2 + by^2)(x+y) = ax^3 + by^3 + ax^2y + bxy^2 \to 217(x+y) = 691 + 79xy$$
hence $x + y = 1, xy = -6$. Now:
$$(ax^3 + by^3)(x+y) = ax^4 + by^4 + ax^3y + bxy^3 = ax^4 + by^4 + xy(ax^2 + by^2)$$
$$691\times 1 = ax^4 + by^4 -6 \times 217$$
thus $ax^4 + by^4 = 1993$.
