Determine $ax^4 + by^4$ for system of equations I found the following recreational problem without further specification for $a,b$. 
Let $x,y$ be real numbers s.t.
$a + b = 6$, 
$ax + by = 10$, 
$ax^2 + by^2 = 24$,
$ax^3 + by^3 = 62$.
Determine $ax^4 + by^4$. 
I am new to problem solving exercises like this and therefore appreciate diverse approaches to this problem as well as comments on how to tackle those types of exercises. 
 A: $$ \color{red}{3 \cdot 62 - 24 = 162}.   $$
$$ 3 \cdot 10 - 6 = 24.   $$
$$ 3 \cdot 24 - 10 = 62.   $$
This is the observation of @Ross,
$$  (xy-1)(x-y)^2 = 0.  $$ Ross points out in comment (below) that an ingredient was $44=abxy(x-y)^2,$ so that
$$ a \neq 0, b \neq 0, x \neq 0, y \neq 0, x \neq y. $$ Finally,
$$  xy = 1.  $$
The simple fact I got was
$$ a (x^2 - 3x+1) + b ( y^2 - 3 y + 1) = 0.  $$
We have $xy = 1.$ If $x^2 - 3 x + 1 \neq 0$ and $y^2 - 3 y + 1 \neq 0,$
$$  y^2 - 3 y + 1 = (x^2 - 3 x + 1) / x^2,$$
$$ x^2 ( y^2 - 3 y + 1 ) = x^2 - 3 x + 1.  $$
$$ a x^2 (y^2 - 3 y + 1) + b (y^2 - 3 y + 1) = 0,$$
$$ (a x^2 + b)(y^2 - 3 y + 1) = 0.  $$
Switch the letters, we get
$$ (a  + b y^2)(x^2 - 3 x + 1) = 0.  $$
Alright,
$$ a x^2 + b y^2 = 24, a x^2 + b = 0.  $$
$$  b(y^2 - 1) = 24. $$ Also
$$  a ( x^2 - 1) = 24. $$
Add,
$$   a(x^2 - 1) + b ( y^2 - 1) = 48.$$ However, we know
$$ a x^2 + b y^2 - a - b = 24 - 6 = 18.  $$ This contradicts the assumption 
$ x^2 - 3 x + 1 \neq 0.   $
$$  \color{red}{ x^2 - 3 x + 1 = 0} $$
$$ \color{red}{  y^2 - 3 y + 1 = 0} $$
A: We can start with $$(ax+by)(ax^3+by^3)-(ax^2+by^2)^2=620-24^2=44=abxy(x-y)^2\\
(a+b)(ax^2+by^2)-(ax+by)^2=44=ab(x-y)^2$$ Which shows $xy=1$  Now, if nothing else, you can rewrite the middle two in terms of $a,x$ and solve them.
A: This is not a nice solution.
Considering the equations $$a+b-6=0\tag 1$$ $$ax+by-10=0\tag 2$$ $$ax^2+by^2-24=0\tag 3$$ $$ax^3+by^3-62=0\tag 4$$ Let us eliminate the variables one at the time and express them as a function of $a$.
From $(1)$, $b=6-a$. Replacing in $(2)$ and assuming $a\neq6$, we have $$a x+(6-a) y-10=0 \implies y=\frac{a x-10}{a-6}\tag 5$$ Replacing $b$ and $y$ in $(3)$, we have $$\frac{a \left(-6 x^2+20 x-24\right)+44}{a-6}=0\implies x_{\pm}=\frac{5a\pm\sqrt{11} \sqrt{6 a-a^2}}{3 a}\tag 6$$ (assuming $a\neq 0$).
Let us use $x_+$ and replace in $(4)$ to end with $$\frac{22 \left(\sqrt{(6-a) a}-2 \sqrt{11} a+6 \sqrt{11}\right)}{9 \sqrt{(6-a) a}}=0 \tag 7$$ So, we need to solve $$\sqrt{(6-a) a}=2 \sqrt{11} a-6 \sqrt{11} \tag 8$$ Squaring both sides $$-a^2+6a=44 a^2-264 a+396\implies 45 a^2-270 a+396=0 \implies a_{\pm}=\frac{1}{5} \left(15\pm\sqrt{5}\right)$$ Considering $a_+$ and going backwards, we should get $$a=3+\frac{1}{\sqrt{5}}\qquad b=3-\frac{1}{\sqrt{5}}\qquad x=\frac{1}{2} \left(3+\sqrt{5}\right)\qquad y=\frac{1}{2} \left(3-\sqrt{5}\right)$$ and then the value of $(a x^n+b y^n)$ for any $n$.
Quite laborious, isn't it ? 
Thanks for the fun.
A: I prefer to add another answer for the most general case.
Consider the four equations which I shall write $$ax^{i-1}+b y^{i-1}=c_i \qquad (i=1,2,3,4)$$ Manipulating them, we can show that $x y=\lambda$ and $a b=\mu$ using $$\lambda=\frac{c_3^2-c_2 c_4}{c_2^2-c_1 c_3}$$
$$\mu=-\frac{\left(c_2^2-c_1 c_3\right)^3}{(4 c_4 c_2^3-3 c_3)^2 c_2^2-6 c_1 c_2
   c_3 c_4+c_1 \left(4 c_3^3+c_1 c_4^2\right)}$$ This makes $$a_{\pm}=\frac{1}{2} \left(c_1\pm\sqrt{c_1^2-4 \mu }\right)$$ Using $a_+$ , this gives $$b=c_1-a\qquad x=\frac{c_2+\sqrt{c_2^2-4 \lambda  \mu }}{2 a}\qquad y=\frac \lambda x$$ Using the given values for the $c_i$'s, this leads to $$\lambda=1\qquad \mu=\frac{44}{5}$$ $$a=3+\frac{1}{\sqrt{5}}\qquad b=3-\frac{1}{\sqrt{5}}\qquad x=\frac{1}{2} \left(3+\sqrt{5}\right)\qquad y=\frac{1}{2} \left(3-\sqrt{5}\right)$$
Now, play with the numbers of your choice.
