1
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Have a look at Newton identities and almost at en.wikipedia.org/wiki/… $\endgroup$ – Claude Leibovici Nov 17 '15 at 12:02
  • $\begingroup$ 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. $\endgroup$ – Abstraction Nov 17 '15 at 12:03
1
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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... $$\sum_{cyc} x_1^3 = \left(\sum_{cyc} x_1 \right)^3-3\left(\sum_{cyc} x_1 \right)\left(\sum_{cyc} x_1 x_2 \right)+3\sum_{cyc} x_1x_2x_3$$

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.