Sum of cubes of roots of a quartic equation $x^4 - 5x^2 + 2x -1= 0$
What is the sum of cube of the roots of equation other than using substitution method? Is there any formula to find the sum of square of roots, sum of cube of roots, and sum of fourth power of roots for quartic equation?

Easy way to do this is to take the first derivative of the polynomial and divide it by the polynomial itself.
f ( x ) = x^4 - 5x^2 + 2x - 1 = 0 
f ' ( x ) = 4x^3 - 10x + 2
Now do f ' ( x ) / f ( x )  =  ?
(x^4 + 0x^3 - 5x^2 + 2x - 1 )   ) ......4x^3 + 0x^2 - 10x + 2
.......- 0....+5........- 2..+ 1
4 / 1 = 4 so that is our first coefficient.
multiply all these numbers - 0.....+ 5....- 2 and + 1 by 4  
Place the answers under the columns to the right. 
so it would look like this :   4x^3 + 0x^2 - 10x + 2
.....................................................- 0..........+20........-8..........+ 1
so the sum of the roots is 0, but the sums of the squares of the roots is +10.
