# 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?

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Cube of root means that if the root is x then cube of the root is x^3. –  Ghanaraj Devarajan Aug 12 '14 at 7:32

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 $j < 0$.

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.

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You should explain to the OP the definition of the notation you are using as I don't believe it is standard. –  Alizter Aug 12 '14 at 7:40
My intent was for the OP to read the link first. My bad. I will edit my answer to make it more self-contained. –  JimmyK4542 Aug 12 '14 at 7:41

we can label the 4 roots $r_1,r_2,r_3,r_4$

The polynomial clearly factors then into:

$$(x-r_1)(x-r_2)(x-r_3)(x-r_4)$$

Given this it becomes evident that:

$$-1 = r_1r_2r_3r_4$$ $$2 = -r_1r_2r_3 -r_1r_2r_4...$$ $$-5 = r_1r_2 + r_1r_3 + r_1r_4...$$ $$0 = -r_1 - r_2 - r_3 -r_4$$

Now using this we can reason about

$$L = (r_1 + r_2 + r_3 + r_4)^3$$

We know that when expanded it will take on the form:

$$L = r_1^3 + ... + r_4^3 + 3r_1^2r_2 + 3r_1^2r_3 + ... 3r_3^2r_4 + 6r_1r_2r_3 + ... + 6r_2r_3r_4$$

Now if we consider the r value combination associated with the coefficient -5 it becomes quickly apparent that

$$3(r_1 + r_2 + r_3 + r_4)(-5) = L - r_1^3... - r_4^3 + 3(r_1r_2r_3 + r_1r_2r_4 + ... + r_2r_3r_4)$$

Recalling an earlier substitution

$$3(r_1 + r_2 + r_3 + r_4)(-5) = L - r_1^3... - r_4^3 + 3(-2)$$

Now observing the sum of the roots is 0 gives us not only the left hand sum of roots is 0 but L itself must also be 0 and therefore

$$0 = 0 - (r_1^3 + ... r_4^3) -6$$

Implying

$$-6 = r_1^3 + r_2^3 + r_3^3 + r_4^3$$

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