# Homework: Sum of the cubed roots of polynomial

Given $7X^4-14X^3-7X+2 = f\in R[X]$, find the sum of the cubed roots.
Let $x_1, x_2, x_3, x_4\in R$ be the roots. Then the polynomial $X^4-2X^3-X+ 2/7$ would have the same roots. If we write the polynomial as $X^4 + a_1X^3 + a_2X^2 +a_3X + a_4$ then per Viete's theorem:

$a_k = (-1)^k\sigma _k(x_1,x_2,x_3,x_4), k\in \{1,2,3,4\}$, where $\sigma _k$ is the $k$-th elementary symmetrical polynomial. Therefore:

$x_1+x_2+x_3+x_4 = 2$
$x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4 = 0\ (*)$
$x_1x_2x_3 +x_1x_2x_4+x_1x_3x_4+x_2x_3x_4 = 1$
$x_1x_2x_3x_4 = 2/7$

Now how to determine the sum of the cubed roots?
$2^3 = 8= (x_1+x_2+x_3+x_4)(x_1+x_2+x_3+x_4)^2 = (x_1+x_2+x_3+x_4)(x_1^2+x_2^2+x_3^2+x_4^2 + 2(*))$

Here's where things go out of hand:
$(x_1+x_2+x_3+x_4)(x_1^2+x_2^2+x_3^2+x_4^2) = (x_1^3 + x_2^3 + x_3^3+x_4^3) + x_1^2(x_2+x_3+x_4)+x_2^2(x_1+x_3+x_4)+x_3^2(x_1+x_2+x_4)+x_4^2(x_1+x_2+x_3) = 8$
What should I do here?

• Have a look at Newton identities and almost at en.wikipedia.org/wiki/… Nov 17, 2015 at 12:02
• Start from $(x_1+x_2+x_3+x_4)^3$, it expands into sum of sum of cubed roots and some expression involving sum of squared roots and known values. Nov 17, 2015 at 12:03

Hint: as we can have only cubic terms in the symmetric polynomial sums, the only terms which can be used are of form $(\sum x)^3, \sum x \sum xy$ and $\sum xyz$. Then it is a matter of testing $3$ coefficients... $$\sum_{cyc} x_1^3 = \left(\sum_{cyc} x_1 \right)^3-3\left(\sum_{cyc} x_1 \right)\left(\sum_{cyc} x_1 x_2 \right)+3\sum_{cyc} x_1x_2x_3$$

Let $$A=x_1+x_2+x_3+x_4=2$$ $$B=x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4=0$$ $$C=x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4=1$$ $$D=x_1x_2x_3x_4=\frac 27.$$ $$E=x_1^2x_2+x_1x_2^2+x_1^2x_3+x_1x_3^2+x_1^2x_4+x_1x_4^2+x_2^2x_3+x_2x_3^2+x_2^2x_4+x_2x_4^2+x_3^2x_4+x_3x_4^2$$

We have $$A^3=x_1^3+x_2^3+x_3^3+x_4^3+3E+6C$$ and $$AB=E+3C.$$

So, $$x_1^3+x_2^3+x_3^3+x_4^3=A^3-3(AB-3C)-6C=\color{red}{11}.$$

Rewrite the equation $7X^4-14X^3-7X+2 = 0$ this way: $7X^3-14X^2-7+\frac{2}{X} = 0$. Replacing $X$ by $X_1$ then $X_2$ etc. and getting the sum will have: $$7\sum{X_i^3} -14\sum{X_i^2}-7\sum{X_i}+2\sum{\frac{1}{X_i}}=0 \tag1$$ From $(1)$ should be easy to get the cube root's sum.

Note that $\sum{\frac{1}{X_i}}$ can be easily calculated using Viete.

Use Newton's relations between sums of powers $p_k=\sim_ix_i^k$ and the elementary symmetric functions: \begin{align*} p_1&=\sigma_1,\\ p_2&=\sigma_1p_1-2\sigma_2,\\ p_3&=\sigma_1p_2-\sigma_2p_1+3\sigma_3. \end{align*}