# Finding $x^8+y^8+z^8$, given $x+y+z=0$, $\;xy +xz+yz=-1$, $\;xyz=-1$

The system says $$x+y+z=0$$ $$xy +xz+yz=-1$$ $$xyz=-1$$ Find $$x^8+y^8+z^8$$

With the first equation I squared and I found that $$x^2+y^2+z^2 =2$$ trying with

$$(x + y + z)^3 = x^3 + y^3 + z^3 + 3 x^2 y + 3 x y^2 + 3 x^2 z++ 3 y^2 z + 3 x z^2 + 3 y z^2 + 6 x y z$$ taking advantage of the fact that there is an $$xyz=-1$$ in the equation, but I'm not getting anywhere, someone less myopic than me.how to solve it?

Thanks

Edit : Will there be any university way to solve this problem , they posed it to a high school friend and told him it was just manipulations of remarkable products. His answers I understand to some extent but I don't think my friend understands all of it.

• Hint: $\,x,y,z\,$ are the roots of $\,t^3-t+1=0\,$ by Vieta's relations, so $\,x^3=x-1\,$ and $\dots$
– dxiv
Sep 9 at 22:34
• Similarly, but alternatively, you could figure out the polynomial (using Vieta) that has $x^2, y^2, z^2$ as roots. If you feed the resulting polynomial into this same method, you'll get a polynomial with $x^4, y^4, z^4$ as roots. Then after one more application of this method... Sep 9 at 22:44
• We can also use Girard -newton identities . Sep 9 at 22:46
• To continue your approach, you may find it handy to expand $(x + y + z)(xy + yz + xz)$. Sep 9 at 23:13
• @Nabla - You are propably aware of this: There is an explicit formula for the set of zeros of $f(t)=t^3-t+1$: You get an explicit formula for $x,y,z$ and may calculate any power $x^n+y^m+z^k$. Sep 10 at 8:23

The following is an elementary, entirely self-contained solution which does not assume knowledge of Vieta's relations, Newton's identities, or anything other than simple algebra. (This is not the best or recommended way to solve it - see the other answers and comments for that - but it answers OP's edit asking for a more basic solution.)

• From $$x+y+z=0$$ it follows that:

$$y+z = -x \tag{1}$$

• From $$xy +xz+yz=-1$$ and $$(1)$$ it follows that: $$-1=x(y+z)+yz=-x^2 + yz \;\;\implies\;\; yz = x^2-1 \tag{2}$$

• From $$xyz=-1$$ and $$(2)$$ it follows that:

$$-1=xyz=x(x^2-1)=x^3-x \;\;\implies\;\; x^3=x-1 \tag{3}$$

Multiplying $$(3)$$ by $$x$$ gives:

$$x^4=x^2-x \tag{4}$$

Squaring $$(4)$$ and using that $$x^3=x-1$$ per $$(3)\,$$:

\begin{align} x^8 &= x^4-2x^3+x^2 \\ &=x \cdot x^3-2x^3+x^2 \\ &=x(x-1)-2(x-1)+x^2 \\ &=2x^2-3x+2 \tag{5} \end{align}

Repeating the steps $$(1)\dots(5)$$ for the other variables, it follows by symmetry that:

\begin{align} y^8 = 2y^2-3y+2 \tag{6} \\ z^8 = 2z^2-3z+2 \tag{7} \end{align}

Adding $$(5)+(6)+(7)\,$$ and using that $$x+y+z=0$$:

\require{cancel} \begin{align} x^8+y^8+z^8 &= 2(x^2+y^2+z^2) - \cancel{3(x+y+z)} + 6 \\ &= 2\left(\bcancel{(x+y+z)^2}-2(xy+yz+zx)\right)+6 \\ &= -4 \cdot (-1) + 6 \\ &= 10 \end{align}

• I like this approach which is simpler and more in spirit of modern algebra ($x$ is a root of a third degree polynomial so any rational function of $x$ can be written as a quadratic polynomial in $x$). +1 Sep 11 at 13:12

If you're not familiar to Girard-Newton identities, a recurrence relation may help:

\begin{align} \qquad a^n+b^n+c^n & = (a+b+c)(a^{n-1}+b^{n-1}+c^{n-1}) \\ & \quad-(bc+ca+ab)(a^{n-2}+b^{n-2}+c^{n-2})+abc(a^{n-3}+b^{n-3}+c^{n-3}) \end{align}

By successive iterations, you can raise up to the desired order.

Using SymmetricReduction[x^8+y^8+z^8,{x,y,z}] in Wolfram Alpha, the result is here.

• Is that supposed to be $$a^n + b^n + c^n = (a + b + c)(a^{n-1} + b^{n-1} + c^{n-1}) - (bc + ca + ab)(a^{n-2} + b^{n-2} + c^{n-2}) + abc(a^{n-3} + b^{n-3} + c^{n-3}),$$ @NgChungTak? Sep 11 at 9:44
• @ArnieBebita-Dris Corrected with thanks. Sep 11 at 10:52

Denoting with $$e_k(x,y,z), k\geq 0$$ the elementary symmetric polynomials \begin{align*} e_1(x,y,z)&=x+y+z=0\\ e_2(x,y,z)&=xy+xz+yz=-1\tag{1}\\ e_3(x,y,z)&=xyz=-1 \end{align*} and with $$p_k(x,y,z), k\geq 0$$ the $$k$$-th power sum \begin{align*} p_k(x,y,z)=x^k+y^k+z^k\tag{2} \end{align*} we recall Newtons identities admit a generating function representation of the power sums as \begin{align*} \color{blue}{\sum_{k=1}^\infty(-1)^{k-1}p_k\frac{t^k}{k}=\ln(1+e_1t+e_2t^2+e_3t^3+\cdots)}\tag{3} \end{align*}

We obtain from (1) - (3) \begin{align*} \color{blue}{\left.x^8+y^8+z^8\right|_{{{e_1=0\ \ }\atop{e_2=-1}}\atop{e_3=-1}}} &=(-8)[t^8]\ln\left(1-t^2-t^3\right)\\ &=(-8)[t^8]\ln\left(1-t^2(1+t)\right)\\ &=8[t^8]\sum_{j=1}^\infty\frac{t^{2j}(1+t)^j}{j}\tag{4.1}\\ &=8\sum_{j=1}^{4}\frac{1}{j}[t^{8-2j}](1+t)^j\tag{4.2}\\ &=8\sum_{j=1}^4\frac{1}{j}\binom{j}{8-2j}\tag{4.3}\\ &=8\left(\frac{1}{3}\binom{3}{2}+\frac{1}{4}\binom{4}{0}\right)\tag{4.4}\\ &\,\,\color{blue}{=10} \end{align*}

Comment:

• In (4.1) we expand the logarithmic series.

• In (4.2) we apply the rule $$[t^{p-q}]A(t)=[t^p]t^qA(t)$$. We also set the upper limit of the series to $$4$$ since other terms do not contribute.

• In (4.3) we select the coefficient of $$t^{8-2j}$$.

• In (4.4) we use $$\binom{p}{q}=0$$ if $$0 are positive integers.

Note that $$x$$, $$y$$, $$z$$ are the roots of the polynomial $$\lambda^3 - \lambda + 1$$, so they are the eigenvalues of the companion matrix $$A=\left(\begin{matrix} 0 & 0 & -1 \\ 1 & 0 & 1\\ 0 & 1 & 0 \end{matrix} \right)$$ Conclude that the eigenvalues $$A^{8}$$ are $$x^8$$, $$y^8$$, $$y^8$$. Calculate $$A^8$$ by squaring three times ( @Theo Bendit:'s idea in the comments). We get $$A^8=\left(\begin{matrix} 2 & -2 & 3 \\ -3 & 4 & -5\\ 2 & -3 & 4 \end{matrix} \right)$$ so the sum $$x^8+y^8+z^8$$ equals the trace of $$A^8$$, that is $$10$$.

• , you can clarify from here " We conclude that the eigenvalues...." Sep 10 at 21:17
• @Nabla: The matrix $A$ can be diagonalized, with $x$, $y$, $z$ on the diagonal. So $A^8$ can also be diagonalized, with $x^8$, $\ldots$ on the diagonal. This is true for companion matrices $A$ of a polynomial with simple roots. Sep 10 at 21:22
• I think it would read better like this: "... $x$, $y$, $z$ are the roots of $\lambda^3 - \lambda + 1$, so they are the eigenvalues of the companion matrix..." Sep 11 at 18:52

$$x, y$$ and $$z$$ are roots of the polynomial $$x^3-x+1=0$$

In the usual notation ,we have $$\Sigma x^2=(\Sigma x)^2-2\Sigma xy=0-2(-1)=2$$ Then since the relation $$x^3=x-1$$ holds for all three roots, summing these gives:

$$\Sigma x^3=\Sigma x - \Sigma 1=0-3=-3$$

Likewise, $$\Sigma x^4=\Sigma x^2-\Sigma x=2$$ $$\Sigma x^5=\Sigma x^3-\Sigma x^2=-3-2=-5$$ $$\Sigma x^6=\Sigma x^4-\Sigma x^3=2--3=5$$ And finally, $$\Sigma x^8=\Sigma x^6-\Sigma x^5=5--5=10$$

• hi , would you be so kind as to clarify a little what you have done? Sep 10 at 21:14
• @ Nabla where are you stuck? Sep 10 at 23:16

This is an expansion of my comment. Recall Vieta's formulas (in this case, for cubics):

If $$p, q, r$$ are the roots of $$t^3 + bt^2 + ct + d$$, where $$a \neq 0$$, then \begin{align*} p + q + r &= -b \\ pq + qr + pr &= c \\ pqr &= -d. \end{align*}

This follows easily by just expanding $$(t - p)(t - q)(t - r)$$ and equating coefficients with $$t^3 + bt^2 + ct + d$$.

Using these formulas, we can see, as others have commented, that the numbers $$x, y, z$$ in your question are just roots of the polynomial $$t^3 - t + 1$$. We can also use Vieta's formulas, and a little computation along the lines of what you've attempted already, to obtain a polynomial whose roots are precisely the squares of the roots of the original polynomial.

Let's say that $$t^3 + bt^2 + ct + d$$ has roots $$p, q, r$$ as above. Let $$t^3 + b't^2 + c't + d'$$ have roots $$p^2, q^2, r^2$$. We wish to compute $$b', c' d'$$ vie Vieta's formulas, i.e. to compute $$p^2 + q^2 + r^2$$, $$p^2 q^2 + q^2 r^2 + p^2 r^2$$, and $$p^2 q^2 r^2$$ in terms of $$a, b, c$$.

There's a quick and easy one: $$p^2 q^2 r^2 = (pqr)^2 = d^2$$. Thus $$d' = -d^2$$.

To compute $$p^2 + q^2 + r^2$$, consider $$b^2 = (p + q + r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + rp) = -b' + 2c,$$ thus $$b' = 2c - b^2$$.

Lastly, we compute \begin{align*} c^2 &= (pq + qr + pr)^2 \\ &= p^2 q^2 + q^2 r^2 + p^2r^2 + 2p^2qr + 2pq^2r + 2pqr^2 \\ &= c' + 2pqr(p + q + r) \\ &= c' + 2bd, \end{align*} so $$c' = c^2 - 2bd$$.

Thus, our transformation which squares roots takes a monic polynomial $$t^3 + bt^2 + ct + d$$ and transforms it into the monic polynomial $$t^3 + (2c - b^2)t^2 + (c^2 - 2bd)t - d^2.$$

Let's apply it to $$t^3 - t + 1$$ to get a polynomial whose roots are $$x^2, y^2, z^2$$. We obtain the polynomial: $$t^3 + (2(-1) - 0^2)t^2 + ((-1)^2 - 2(0)(1))t - 1^2 = t^3 - 2t^2 + t - 1.$$ Apply again to get a polynomial whose roots are $$x^4, y^4, z^4$$: $$t^3 + (2(1) - (-2)^2)t^2 + (1^2 - 2(-2)(-1))t - (-1)^2 = t^3 - 2t^2 - 3t - 1.$$ We could apply this transformation one final time to get the polynomial whose roots are $$x^8, y^8, z^8$$, but really we just need the coefficient of $$t^2$$. So, partially applying the above formula, we get: $$t^3 + (2(-3) - (-2)^2)t^2 + \underline{\hspace{12pt}} t + \underline{\hspace{12pt}} = t^3 - 10t^2 + \underline{\hspace{12pt}} t + \underline{\hspace{12pt}}.$$ So, by Vieta, $$x^8 + y^8 + z^8 = -(-10) = 10.$$

You got already that $$x^2+y^2+z^2=2.$$ Also, we have: $$\sum_{cyc}x^2y^2=\left(\sum_{cyc}xy\right)^2-2xyz(x+y+z)=1$$ and $$x^2y^2z^2=1.$$ Thus, $$\sum_{cyc}x^8=\left(\sum_{cyc}x^4\right)^2-2\sum_{cyc}x^4y^4=$$ $$=\left(\left(\sum_{cyc}x^2\right)^2-2\sum_{cyc}x^2y^2\right)^2-2\left(\left(\sum_{cyc}x^2y^2\right)^2-2x^2y^2z^2\sum_{cyc}x^2\right)=$$ $$=(4-2)^2-2(1-2\cdot2)=10.$$

Question: "@hm2020 ,hi , I can't relate the equation you are proposing to the problem, could you clarify? – Nabla"

Answer: There is an explicit formula (you find it on wikipedia under the name "the Cardano formula", or in algebra books) for the roots of $$f(t):=t^3-t+1$$: Using the formula it follows the following number is a root:

$$x:=\sqrt[3]{-\frac{1}{2}+\sqrt{\frac{23}{108}}}+\sqrt[3]{-\frac{1}{2}-\sqrt{\frac{23}{108}} }$$

It follows you may (by hand, in principle) use polynomial division to calculate a decomposition $$f(t)=(t-x)p(t)$$ where $$p(t)$$ is a degree 2 polynomial and there is an explicit formula for the roots of $$p(t)$$. Hence you may give explicit formulas for all the roots $$x,y,z$$ of $$f(t)$$ and also calculate all powers $$x^m+y^n+z^k$$ for all integers $$m,n,k$$. This is a "brute force" method but it still gives explicit formulas. You get the following:

$$f(t)=(t-x)(t^3+xt+(x^2-1))=(t-x)p(t)$$

where $$p(t)=t^2+xt+(x^2-1)$$. The polynomial $$p(t)$$ has roots

$$y:= \frac{1}{2}(-x +\sqrt{4-3x^2})\text{ and } z:=\frac{1}{2}(-x-\sqrt{4-3x^2}).$$

The above formulas give all roots $$x,y,z$$ of $$f(t)$$ and you get

$$F1\text{ }x^m+y^n+z^k:= (\sqrt[3]{-\frac{1}{2}+\sqrt{\frac{23}{108}}}+\sqrt[3]{-\frac{1}{2}-\sqrt{\frac{23}{108}} })^m+$$

$$(\frac{1}{2}(-x +\sqrt{4-3x^2}))^n+(\frac{1}{2}(-x-\sqrt{4-3x^2}))^k.$$

The formula $$F1$$ gives an explicit formula for all $$m,n,k$$, in particular for $$m=n=k=8$$.

You must check the "Cardano formula": It says that a solution to $$g(t):=t^3+pt+q=0$$ is the following:

$$x:=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}} }.$$

The above formulas give all solutions to the equation $$g(t)=0$$. The above calculation give explicit formulas for numbers $$x(p,q), y(p,q), z(p,q)$$ depending on $$p,q$$ solving $$g(t)=0$$.

• @Nabla - you should check that the formula $F1$ agrees with the other formulas given above. Sep 11 at 10:08
• @Nabla - if the above result saying that $x^8+y^8+z^8=10$ is correct, you get the same result with $F1$ - this you must check. Sep 11 at 10:36
• Thank you all very much, even though I have a little knowledge in higher algebra I must admit that some things at the beginning were out of my hands, after the explanations I was much clearer, I did not imagine that the system had such a variety of answers for the same reason I raised it in a general way, expecting a typical substitution here and some other algebraic manipulation there. Then I was forced to edit and ask for a simpler solution. You learn a lot in passing, much appreciated. Sep 12 at 2:22
• @Nabla -the above method is a "brute force" method, still it gives "all solutions". Sep 12 at 11:28
• @Nabla - You should check the "Cardano formula" in a book - wikipedia tends not to be reliable. Sep 12 at 12:33