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.

You could start with $a+b$ and $ab$,  


*

*Note first $(a+b)^2 - 2ab = a^2+ b^2 $

*Note that $(a^2+b^2)^2 - 2(ab)^2 = a^4 + b^4$
As an asside, i use this set of numbers to factorise $b^n-a^n$.  Part of the process involves creating a sequence $T_n = a^n+b^n$.  You find that $T_{n+1}=(a+b)T_n - ab T_{n-1}$
Applying $T_0=2$ and $T_1$=5  we get this series for the sum of $a^n+b^n$ for n=0 t0 6.  The iteration is t(n+1)=5t(n)-2t(n-1).
      0    1    2    3    4     5      6
      2    5   21   95  433  2070   9484

Doing it this way, allows one to evaluate symmetric equations (ie $f(a,b)=f(b,a)), very quickly.  You start at the middle, and work outwards, multiplying the result by ab, before adding the next term.  
