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Solve the system $$\begin{equation} \label{equation1} \begin{split} x+y+z=0 \\ x^3+y^3+z^3=18 \\ x^7+y^7+z^7=2058 \end{split} \end{equation}$$

My work:

Let $$f(p)=p^3+ap^2+bp+c$$ having roots $x,y,z$

By viete's relations we can find that $x+y+z=-a$ or $z+x+y=0$ So our polynomial becomes $$f(p)=p^3+bp+c$$ Since $x,y,z$ are roots $$\begin{equation} \label{equation2} \begin{split} x^3+bx+c=0 \\ y^3+by+c=0 \\ z^3+bz+c=0 \end{split} \end{equation}$$ Adding all equations $$x^3+y^3+z^3+b(x+y+z)+3c=0$$ or $$18+b\cdot0+3c=0$$ or $$c=-6$$ Our polynomial becomes $$f(p)=p^3+bp-6$$ After this I'm stuck. I can't find the value of $b$. Any help is greatly appreciated.

Blue
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abcdefu
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2 Answers2

3

There is always the way of a direct computation. It works the same way as here:

Solving $x+y+z=4$, $x^2+y^2+z^2=14$, $x^3+y^3+z^3=34$

Substituting $z=-x-y$, the other two equations are $f(x,y)=0$ and $g(x,y)=0$, and the resultant of $f$ and $g$ with respect to $y$ yields $$ x(x^3+7x-6)(x+2)(x+1)(x-3)=0. $$

Dietrich Burde
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0

A bit late answer but I think worth mentioning it.

Here is a solution based on linear recurrences.

If $x,y,z$ are different zeros of your polynomial $f(p) = p^3 +bx -6$, then for natural $n\geq 0$

$$a_n = x^n+y^n+z^n$$

is the solution of the recurrence

$$a_{n+3} = -b\cdot a_{n+1} + 6a_n$$

with \begin{eqnarray*} a_0 & = & x^0+y^0+z^0 & = & 3 \\ a_1 & = & x^1+y^1+z^1 & = & 0 \\ a_3 & = & x^3+y^3+z^3 & = & 18 \\ \ldots \end{eqnarray*}

Now, expanding $(x+y+z)^2$ you find $$-b=-(xy+yz+zx) = \frac 12a_2$$

Using this together with the recurrence you get:

\begin{eqnarray*} a_4 & = & \frac 12 a_2a_2 + 6a_1 & \stackrel{a_1=0}{=} &\frac 12 a_2^2\\ a_5 & = & \frac 12 a_2 a_3 + 6a_2 & = & 15a_2 \\ 2058 & = & \frac 12 a_2 a_5 +6a_4& = & \frac{21}2 a_2^2 \\ \ldots \end{eqnarray*}

It follows $$a_2^2=196 \stackrel{a_2 > 0}{\Rightarrow} a_2 = 14 \Rightarrow b= -7$$

So, you get

$$f(p) = p^3-7p-6 \text{ with the easy to find zeros } -1,-2,3.$$

trancelocation
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