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?
This is precisely what Newton's Sums are for. 

Let $S_n$ be the sum of the $n$-th powers of the roots of $a_nx^n+a_{n-1}x^{n-1} + \cdots + a_1x+a_0$. 

Then, Newton's Sums tells us that $\displaystyle\sum_{k = 0}^{m-1}\left(a_{n-j}S_{m-j}\right)+ma_{n-m} = 0$ for each positive integer $m$.  

We can write out the first few equations explicitly as follows: 

$a_nS_1 + a_{n-1} = 0$

$a_nS_2 + a_{n-1}S_1 + 2a_{n-2} = 0$

$a_nS_3 + a_{n-1}S_2 + a_{n-2}S_1 + 3a_{n-3} = 0$

$a_nS_4 + a_{n-1}S_3 + a_{n-2}S_2 + a_{n-3}S_1 + 4a_{n-4} = 0$

and so on. Note that we define $a_j = 0$ for $j &lt; 0$. 

Applying those formula to the polynomial $1x^4+0x^3-5x^2+2x+1$ yields the following:

$1 \cdot S_1 + 0 = 0 \leadsto S_1 = 0$ (sum of roots)

$1 \cdot S_2 + 0 \cdot S_1 + 2 \cdot -5 = 0 \leadsto S_2 = 10$ (sum of squares of roots)

$1 \cdot S_3 + 0 \cdot S_2 + - 5 \cdot S_1 + 3 \cdot 2 = 0 \leadsto S_3 = -6$ (sum of cubes of roots)

and so on.
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.