<s>[INST] 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.
 [/INST] Hint:
$$(a+b)(ax^2+by^2)-(ax+by)^2=\cdots=ab(x-y)^2\ \ \ \ (1)$$
$$(a+b)(ax^3+by^3)-(ax+by)(ax^2+by^2)=\cdots=ab(x-y)^2(x+y)\ \ \ \ (2)$$
On division, we get $x+y$
$$(a+b)(ax^3+by^3)-(ax^2+by^2)^2=\cdots=abxy(x-y)^2\ \ \ \ (3)$$
$(1)/(3)$ will give $xy$
$$(a+b)(ax^4+by^4)-(ax+by)(ax^3+by^3)=\cdots=ab(x-y)^2((x+y)^2-xy)\ \ \ \ (4)$$
$(4)/(3)$ will give $$\dfrac{23(ax^4+by^4)-79\cdot691}{23(691)-217^2}=\dfrac{xy}{(x+y)^2-xy}$$
Now replace the values of $x+y$ and $xy$
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