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How to show that they are equal? All I can come up with is using symmetric polynomials to express them, or using some substitution to simplify this identity since it is symmetric and homogeneous but they are still too complicated for one to work out during the exam. So I think there should exist some better approaches to handle this identity without too much direct computation.

In addition, this identity is supposed to be true:

$$ \frac{x^7+y^7+z^7}{7}=\frac{x^2+y^2+z^2}{2}\times\frac{x^5+y^5+z^5}{5} .$$

amWhy
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Bach
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4 Answers4

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Write $S_n=x^n+y^n+z^n$. Then $S_0=3$, you are given $S_1=0$ and are charged to prove that $S_5=(5/6)S_2S_3$. Define $$F(t)=\sum_{n=0}^\infty S_nt^n.$$ Then $$ F(t)=\frac1{1-xt} + \frac1{1-yt} + \frac1{1-zt} = \frac{3+2e_1t+e_2t^2} {1-e_1t+e_2t^2-e_3t^3} $$ where $e_1=x+y+z$, $e_2=xy+xz+yz$ and $e_3=xyz$. Then $e_1=0$ so \begin{align} F(t)&=\frac{3+e_2t^2} {1+e_2t^2-e_3t^3}=(3+e_2t^2)\sum_{k=0}^\infty(-1)^k(e_2t^2-e_3t^3)^k\\ &=(3+e_2t^2)(1-e_2t^2+e_3t^3+e_2^2t^4-2e_2e_3t^5+\cdots)\\ &=3-2e_2t^2+3e_3t^3+2e_2^2t^4-5e_2e_3t^5+\cdots. \end{align} Therefore $S_2=-2e_2$, $S_3=3e_3$ and $S_5=-5e_2e_3$ etc.

Bach
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Angina Seng
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First we can use the identities

$$ (x+y+z)^2=x^2+y^2+z^2+2(yz+zx+xy) $$

and

$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-yz-zx-xy)$$

As $x+y+z=0$, we have $x^2+y^2+z^2=-2(yz+zx+xy)$ and $x^3+y^3+z^3=3xyz$.

So, R.H.S is $-xyz(yz+zx+xy)$.

From the above results, we have

\begin{align*} (x^3+y^3+z^3)(x^2+y^2+z^2)&=x^5+y^5+z^5+x^3(y^2+z^2)+y^3(z^2+x^2)+z^3(x^2+y^2)\\ (3xyz)(-2yz-2zx-2xy)&=x^5+y^5+z^5+x^3[(y+z)^2-2yz]\\ &\quad +y^3[(z+x)^2-2zx]+z^3[(x+y)^2-2xy]\\ -6xyz(yz+zx+xy)&=x^5+y^5+z^5+x^5-2x^3yz+y^5-2xy^3z+z^5-2xyz^3\\ -6xyz(yz+zx+xy)&=2(x^5+y^5+z^5)-2xyz(x^2+y^2+z^2)\\ -6xyz(yz+zx+xy)&=2(x^5+y^5+z^5)+4xyz(yz+zx+xy)\\ x^5+y^5+z^5&=-5xyz(yz+zx+xy) \end{align*}

Hence, L.H.S. is also $-xyz(yz+zx+xy)$.

Bach
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CY Aries
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By Newton's identities, with $e_k$ standing for the elementary symmetric polynomials and $p_k$ for the sum of $k^{th}$ powers, and using that $e_1=0\,$:

$$ \begin{align} p_1 &= e_1 &&= 0\\ p_2 &= e_1p_1-2e_2 &&= -2e_2 \\ p_3 &= e_1p_2 -e_2p_1 + 3e_3 &&= 3e_3 \\ p_4 &= e_1p_3 - e_2p_2+e_3p_1-4e_4 &&= 2 e_2^2 \\ p_5 &= e_1p_4-e_2p_3+e_3p_2-e_4p_1+5e_5 &&= -5e_2e_3 \end{align} $$

It follows that:

$$\color{blue}{\frac{p_5}{5}} = -e_2e_3 = \frac{-2e_2}{2} \cdot \frac{3e_3}{3} = \color{blue}{\frac{p_2}{2} \cdot \frac{p_3}{3}} $$

dxiv
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  • Hhhh, nice! I think this is the method I am thinking about! But if one forget what the exact formulas of these identities(probably most of the time, hhhh) are, it would be annoying to recover each term by oneself. – Bach Apr 21 '18 at 19:07
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    @Yuchen If you remember the general form of the identity (also listed at the posted link) then, even if you forget the details, you can fairly easily write down $p_1,p_2,p_3$ then you should recognize the pattern. – dxiv Apr 21 '18 at 19:09
  • Indeed! Thank you~ – Bach Apr 21 '18 at 19:23
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If $x=0$, the equality is trivial. So, it is sufficient to consider $x\ne 0$. Let $y=ax, z=abx$. Then: $$x+ax+abx=0 \Rightarrow ab=-1-a.$$ Then: $$\frac{x^5+a^5x^5+(ab)^5x^5}{5}=\frac{x^2+a^2x^2+(ab)^2x^2}{2}\cdot \frac{x^3+a^3x^3+(ab)^3x^3}{3} \Rightarrow$$ $$\frac{1+a^5+(-1-a)^5}{5}=\frac{1+a^2+(-1-a)^2}{2}\cdot \frac{1+a^3x^3+(-1-a)^3}{3} \Rightarrow$$ $$-a-2a^2-2a^3-a^4=(1+a+a^2)(-a-a^2).$$

farruhota
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