System of equations involving 4 variables 
If $$a + b = 6$$
$$ax + by = 10$$
$$ax^2 + by^2 = 24$$
$$ax^3 + by^3 = 62$$
then $$ax^4 + by^4 = ?$$

I got $$a(x-1) + b(y-1) = 4$$ $$ax(x-1) + by(y-1) = 14$$ $$ax^2(x-1) + by^2(y-1) = 38$$ by subtracting the given equations and also got $$a(x-1)^2 + b(y-1)^2 = 10$$ $$ax^2(x-1)^2 + by^2(y-1)^2 = 24$$ by further subtracting the three equations.
I don't think I'm going anywhere with this process and so I'm not sure how to approach this. Any help is appreciated.

Define $A_i$ as follows
\begin{eqnarray*}
a+b=A_0 \\ 
ax+by=A_1 \\ 
ax^2+by^2=A_2 \\
ax^3+by^3=A_3 \\
ax^4+by^4=A_4 .
\end{eqnarray*}
Multiply the first & third equations & subtract the square of the second equation
\begin{eqnarray*}
A_0 A_2 -A_1^2 = ab(x-y)^2.
\end{eqnarray*}
Multiply the first & fourth equations & subtract the product of the second & third equations
\begin{eqnarray*}
A_0 A_3 -A_1 A_2 = ab(x+y)(x-y)^2 \\
(x+y) = \frac{A_0 A_3 -A_1 A_2}{A_0 A_2 -A_1^2}.
\end{eqnarray*}
Now multiply the first & fifth equations & subtract the square of the third equation
\begin{eqnarray*}
A_0 A_4 - A_2^2 = ab(x^2-y^2)^2 = ab(x+y)^2(x-y)^2 = \frac{(A_0 A_3 -A_1 A_2)^2}{A_0 A_2 -A_1^2}.
\end{eqnarray*}
Now plug in your values of $A_0,A_1,A_2,A_3$ and solve for $A_4$.
