Working out the value of $a^4+b^4$ If $ab = 2$ and $a+b = 5$ then calculate the value of $a^4+b^4$
My approach:
$$a^4+b^4 = (a+b)^4-4a^3b-6a^2b^2-4ab^3$$
$$=(5)^4-6(ab)^2-4ab.a^2-4ab.b^2$$
$$=(5)^4-6(24)-4ab(a^2-b^2)$$
$$=(5)^4-6(24)-8(a+b)(a-b)$$
$$=(5)^4-6(24)-8(5)(a-b)$$
I am a little stuck now and any help will be appreciated.

There is a general recursive identity for such questions:  let $f(n) = a^n + b^n$.  Then $$f(n+1) = f(1)f(n) - ab f(n-1).$$  Note that $f(0) = 2$ provided $ab \ne 0$.  Then in your particular case, we wish to find $f(4)$, where $$f(n+1) = 5f(n) - 2f(n-1).$$  With starting values $f(0) = 2$ and $f(1) = 5$, we easily compute $f(2) = 5(5)-2(2) = 21$, $f(3) = 5(21) - 2(5) = 95$, $f(4) = 5(95) - 2(21) = 433$.  The advantage of this approach is that it can be used to compute sums of higher powers quite easily.  It also leads to a general solution via the solution of the associated linear recurrence; e.g., with generating functions.  
